Chemical Kienetics
Topics to be covered
Algorithm to find integrated rate law
Zero Order Reaction
Integrated rate law derivation
Alternative to find integrated rate law
Zero order graph
Example of Zero order reaction
First order reaction
Derivation of integrated first rate law
Alternative to find integrated rate law
First order graph
Gas phase first order reaction
Examples of first order reaction
एकीकृत दर कानून खोजने के लिए एल्गोरिदम
शून्य आदेश प्रतिक्रिया
एकीकृत दर कानून व्युत्पत्ति
एकीकृत दर कानून खोजने का विकल्प
शून्य क्रम ग्राफ
शून्य कोटि अभिक्रिया का उदाहरण
प्रथम कोटि की प्रतिक्रिया
एकीकृत प्रथम श्रेणी कानून की व्युत्पत्ति
एकीकृत दर कानून खोजने का विकल्प
प्रथम क्रम ग्राफ
गैस चरण प्रथम क्रम प्रतिक्रिया
प्रथम कोटि अभिक्रिया के उदाहरण
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Algorithm to find integrated rate law
Write the differential form of rate law.
Rearrange the equation such that the one with the same coeffecient are on the same side.
Integrate both sides.
Put the proper limits to get the final equation.
दर कानून का विभाजनी रूप लिखें।
समीकरण को पुन: व्यवस्थित करें ताकि वही सहसंख्यक वाले एक ही ओर हों।
दोनों ओरों का समाकलन करें।
अंतिम समीकरण प्राप्त करने के लिए उचित सीमाएं लगाएं।
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Zero Order Reaction
Zero order reaction means that the rate of the reaction is proportional to zero power of the concentration of reactants.
Consider the reaction,
R → P Rate = − d [ R ] d t = k [ R ] 0 \small{
\begin{aligned}
& R \to P \\\
& \text{ Rate }=-\frac{d[R]}{d t}=k[R]^0
\end{aligned}
} R → P Rate = − d t d [ R ] = k [ R ] 0
शून्य क्रम की प्रतिक्रिया का अर्थ है कि प्रतिक्रिया की दर प्रतिक्रियाशीलों के संघटक की शून्य शक्ति के अनुपात में होती है।
प्रतिक्रिया पर विचार करें,
R → P दर = − d [ R ] d t = k [ R ] 0 \small{
\begin{aligned}
& R \to P \\\
& \text{ दर }=-\frac{d[R]}{d t}=k[R]^0
\end{aligned}
} R → P दर = − d t d [ R ] = k [ R ] 0
Algorithm to find integrated rate law → \rarr → Zero Order Reaction → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Integrated rate law derivation(I/II)
For a Zero order reaction:
Rate = − d [ R ] d t = k [ R ] 0 \small{
\text{ Rate }=-\frac{d[R]}{d t}=k[R]^0
} Rate = − d t d [ R ] = k [ R ] 0
As any quantity raised to power zero is unity
Rate = − d [ R ] d t = k × 1 d [ R ] = − k d t \small{
\begin{aligned}
& \text{ Rate }=-\frac{d[R]}{d t}=k \times 1 \\\
& d[R]=-k d t
\end{aligned}
} Rate = − d t d [ R ] = k × 1 d [ R ] = − k d t
शून्य क्रम की प्रतिक्रिया के लिए:
दर = − d [ R ] d t = k [ R ] 0 \small{
\text{ दर }=-\frac{d[R]}{d t}=k[R]^0
} दर = − d t d [ R ] = k [ R ] 0
जैसा कि किसी भी मात्रा को शून्य की शक्ति देने से एकता होती है
दर = − d [ R ] d t = k × 1 d [ R ] = − k d t \small{
\begin{aligned}
& \text{ दर }=-\frac{d[R]}{d t}=k \times 1 \\\
& d[R]=-k d t
\end{aligned}
} दर = − d t d [ R ] = k × 1 d [ R ] = − k d t
Algorithm to find integrated rate law → \rarr → → \rarr → Integrated rate law derivation → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Integrated rate law derivation(I/II)
As any quantity raised to power zero is unity
Rate = − d [ R ] d t = k × 1 d [ R ] = − k d t \small{
\begin{aligned}
& \text{ Rate }=-\frac{d[R]}{d t}=k \times 1 \\\
& d[R]=-k d t
\end{aligned}
} Rate = − d t d [ R ] = k × 1 d [ R ] = − k d t
जैसा कि किसी भी मात्रा को शून्य की शक्ति देने से एकता होती है
दर = − d [ R ] d t = k × 1 d [ R ] = − k d t \small{
\begin{aligned}
& \text{ दर }=-\frac{d[R]}{d t}=k \times 1 \\\
& d[R]=-k d t
\end{aligned}
} दर = − d t d [ R ] = k × 1 d [ R ] = − k d t
Algorithm to find integrated rate law → \rarr → → \rarr → Integrated rate law derivation → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Integrated rate law derivation(II/II)
Integrating both sides
[ R ] = − k t + I \small{[R]=-k t+I} [ R ] = − k t + I
where, I is the constant of integration.
At t = 0 \small{t=0} t = 0 R = [ R ] 0 \small{R=[R]_0} R = [ R ] 0 [ R ] 0 \small{[R]_0} [ R ] 0
Substituting in equation
[ R ] 0 = − k × 0 + I [ R ] 0 = I \small{
\begin{aligned}
& {[R]_0=-k \times 0+I} \\\
& {[R]_0=I}
\end{aligned}
} [ R ] 0 = − k × 0 + I [ R ] 0 = I
दोनों ओरों का समाकलन करते हैं
[ R ] = − k t + I \small{[R]=-k t+I} [ R ] = − k t + I
जहाँ, I समाकलन का स्थिरांक है।
t = 0 \small{t=0} t = 0 R = [ R ] 0 \small{R=[R]_0} R = [ R ] 0 [ R ] 0 \small{[R]_0} [ R ] 0
समीकरण में प्रतिस्थापन करते हैं
[ R ] 0 = − k × 0 + I [ R ] 0 = I \small{
\begin{aligned}
& {[R]_0=-k \times 0+I} \\\
& {[R]_0=I}
\end{aligned}
} [ R ] 0 = − k × 0 + I [ R ] 0 = I
Algorithm to find integrated rate law → \rarr → → \rarr → Integrated rate law derivation → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Integrated rate law derivation(II/II)
Substituting the value of I in the equation
[ R ] = − k t + [ R ] 0 \small{
[R]=-k t+[R]_0
} [ R ] = − k t + [ R ] 0
समीकरण में I का मान प्रतिस्थापित करते हैं
[ R ] = − k t + [ R ] 0 \small{
[R]=-k t+[R]_0
} [ R ] = − k t + [ R ] 0
Algorithm to find integrated rate law → \rarr → → \rarr → Integrated rate law derivation → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Alternative to find integrated rate law
∫ [ A ] 0 [ A ] t d [ A ] = ∫ t = 0 t = 1 k d t ⇒ ∫ ∣ A ∣ 0 [ A ] t d [ A ] = k ∫ 0 t d t ⇒ [ A ] [ A ] 0 . [ A ] t = k [ t ] 0 t ⇒ ( [ A ] t − [ A ] 0 ) = k t ∴ [ A ] 0 − [ A ] t = k t ∴ k = [ A ] 0 − [ A ] t t \small{\begin{aligned} & \int _{[A]_0}^{[A]_t} d[A]=\int _{t=0}^{t=1} k d t \\ & ⇒ \int _{|A|_0}^{[A]_t} d[A]=k \int_0^t d t \\ & ⇒ [A] _{[A]_0.}^{[A]_t}=k[t]_0^t \\ & ⇒ ([A]_t-[A]_0)=k t \\ & \therefore[A]_0- [A]_t=k t \\ & \therefore k=\frac{[A]_0- [A]_t}{t} \\ & \end{aligned}} ∫ [ A ] 0 [ A ] t d [ A ] = ∫ t = 0 t = 1 k d t ⇒ ∫ ∣ A ∣ 0 [ A ] t d [ A ] = k ∫ 0 t d t ⇒ [ A ] [ A ] 0 . [ A ] t = k [ t ] 0 t ⇒ ([ A ] t − [ A ] 0 ) = k t ∴ [ A ] 0 − [ A ] t = k t ∴ k = t [ A ] 0 − [ A ] t
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → Alternative to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Zero order graph
Zero order Integerated rate equation
[ R ] = − k t + [ R ] 0 \small{
[R]=-k t+[R]_0
} [ R ] = − k t + [ R ] 0
Comparing it with the equation of straight line:
y = m x + c \small{y = mx + c} y = m x + c
we get,
s l o p e = − k \small{slope = -k} s l o p e = − k
x-axis = t \small{\text{x-axis} = t} x-axis = t
y-axis = concentration at time t \small{\text{y-axis} = \text{concentration at time t}} y-axis = concentration at time t
शून्य क्रम एकीकृत दर समीकरण
[ R ] = − k t + [ R ] 0 \small{
[R]=-k t+[R]_0
} [ R ] = − k t + [ R ] 0
इसे सीधी रेखा की समीकरण के साथ तुलना करते हैं:
y = m x + c \small{y = mx + c} y = m x + c
हमें मिलता है,
ढलान = − k \small{ढलान = -k} ढलान = − k
x-अक्ष = t \small{\text{x-अक्ष} = t} x- अक्ष = t
y-अक्ष = समय t पर संचय \small{\text{y-अक्ष} = \text{समय t पर संचय}} y- अक्ष = समय t पर संचय
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → Zero order graph → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Example of Zero order reaction
Zero order reactions are relatively uncommon but they occur under special conditions.
Some enzyme catalysed reactions and reactions which occur on metal surfaces are a few examples of zero order reactions.
The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at high pressure
2 N H 3 ( g ) → Pt catalyst 1130 K N 2 ( g ) + 3 H 2 ( g ) \small{
2 NH_3(g) \xrightarrow[\text{ Pt catalyst }]{1130 K} N_2(g)+3 H_2(g)
} 2 N H 3 ( g ) 1130 K Pt catalyst N 2 ( g ) + 3 H 2 ( g )
Rate = k [ N H 3 ] 0 = k \small{
\text{ Rate }=k[NH_3]^0=k
} Rate = k [ N H 3 ] 0 = k
शून्य क्रम की प्रतिक्रियाएं अपेक्षाकृत असामान्य होती हैं लेकिन वे विशेष स्थितियों में होती हैं।
कुछ एनजाइम ड्राइवन प्रतिक्रियाएं और धातु की सतह पर होने वाली प्रतिक्रियाएं शून्य क्रम की प्रतिक्रियाओं के कुछ उदाहरण हैं।
गैसीय अमोनिया का विघटन एक गर्म प्लैटिनम सतह पर उच्च दबाव में शून्य क्रम की प्रतिक्रिया है
2 N H 3 ( g ) → Pt catalyst 1130 K N 2 ( g ) + 3 H 2 ( g ) \small{
2 NH_3(g) \xrightarrow[\text{ Pt catalyst }]{1130 K} N_2(g)+3 H_2(g)
} 2 N H 3 ( g ) 1130 K Pt catalyst N 2 ( g ) + 3 H 2 ( g )
Rate = k [ N H 3 ] 0 = k \small{
\text{ Rate }=k[NH_3]^0=k
} Rate = k [ N H 3 ] 0 = k
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Example of Zero order reaction → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Example of Zero order reaction
In this reaction, platinum metal acts as a catalyst.
At high pressure, the metal surface gets saturated with gas molecules.
So, a further change in reaction conditions is unable to alter the amount of ammonia on the surface of the catalyst making rate of the reaction independent of its concentration.
The thermal decomposition of HI on gold surface is another example of zero order reaction.
इस प्रतिक्रिया में, प्लैटिनम धातु एक कैटलिस्ट के रूप में कार्य करती है।
उच्च दबाव पर, धातु की सतह गैस के अणुओं से संतृप्त हो जाती है।
इसलिए, प्रतिक्रिया की स्थितियों में आगे की कोई भी परिवर्तन कैटलिस्ट की सतह पर अमोनिया की मात्रा को बदलने में सक्षम नहीं होता है जिससे प्रतिक्रिया की दर उसकी संघनन से स्वतंत्र हो जाती है।
HI का तापीय विघटन सोने की सतह पर शून्य क्रम की प्रतिक्रिया का एक और उदाहरण है।
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Example of Zero order reaction → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
First order reaction
The rate of the reaction is proportional to the first power of the concentration of the reactant R.
Consider the reaction,
R → P Rate = − d [ R ] d t = k [ R ] or d [ R ] [ R ] = − k d t
\begin{aligned}
& R \to P \\\
& \text{ Rate }=-\frac{d[R]}{d t}=k[R] \\\
& \text{ or } \frac{d[R]}{[R]}=-k d t
\end{aligned}
R → P Rate = − d t d [ R ] = k [ R ] or [ R ] d [ R ] = − k d t
प्रतिक्रिया की दर प्रतिक्रियाशील R की संघटन की पहली शक्ति के अनुपातिक होती है।
प्रतिक्रिया पर विचार करें,
R → P दर = − d [ R ] d t = k [ R ] या d [ R ] [ R ] = − k d t
\begin{aligned}
& R \to P \\\
& \text{ दर }=-\frac{d[R]}{d t}=k[R] \\\
& \text{ या } \frac{d[R]}{[R]}=-k d t
\end{aligned}
R → P दर = − d t d [ R ] = k [ R ] या [ R ] d [ R ] = − k d t
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → First order reaction → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Derivation of integrated first rate law(I/II)
Consider the reaction,
R → P Rate = − d [ R ] d t = k [ R ] or d [ R ] [ R ] = − k d t \small{
\begin{aligned}
& R \to P \\\
& \text{ Rate }=-\frac{d[R]}{d t}=k[R] \\\
& \text{ or } \frac{d[R]}{[R]}=-k d t
\end{aligned}
} R → P Rate = − d t d [ R ] = k [ R ] or [ R ] d [ R ] = − k d t
Integrating this equation, we get
ln [ R ] = − k t + I \small{
\ln [R]=-k t+I
} ln [ R ] = − k t + I
Again, I is the constant of integration and its value can be determined easily.
विचार करें, अभिक्रिया,
R → P दर = − d [ R ] d t = k [ R ] या d [ R ] [ R ] = − k d t \small{
\begin{aligned}
& R \to P \\\
& \text{ दर }=-\frac{d[R]}{d t}=k[R] \\\
& \text{ या } \frac{d[R]}{[R]}=-k d t
\end{aligned}
} R → P दर = − d t d [ R ] = k [ R ] या [ R ] d [ R ] = − k d t
इस समीकरण का समाकलन करते हैं, हमें मिलता है
ln [ R ] = − k t + I \small{
\ln [R]=-k t+I
} ln [ R ] = − k t + I
फिर से, I समाकलन का स्थिरांक है और इसका मान आसानी से निर्धारित किया जा सकता है।
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Derivation of integrated first rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Derivation of integrated first rate law(I/III)
When t = 0 , R = [ R ] 0 \small{t=0, R=[R]_0} t = 0 , R = [ R ] 0 [ R ] 0 \small{[R]_0} [ R ] 0
ln [ R ] 0 = − k × 0 + I ln [ R ] 0 = I \small{
\begin{aligned}
& \ln [R]_0=-k \times 0+I \\\
& \ln [R]_0=I
\end{aligned}
} ln [ R ] 0 = − k × 0 + I ln [ R ] 0 = I
Substituting the value of I in equation
ln [ R ] = − k t + ln [ R ] 0 \small{
\ln [R]=-k t+\ln [R]_0
} ln [ R ] = − k t + ln [ R ] 0
जब t = 0 , R = [ R ] 0 \small{t=0, R=[R]_0} t = 0 , R = [ R ] 0 [ R ] 0 \small{[R]_0} [ R ] 0
ln [ R ] 0 = − k × 0 + I ln [ R ] 0 = I \small{
\begin{aligned}
& \ln [R]_0=-k \times 0+I \\\
& \ln [R]_0=I
\end{aligned}
} ln [ R ] 0 = − k × 0 + I ln [ R ] 0 = I
समीकरण में I का मान प्रतिस्थापित करना
ln [ R ] = − k t + ln [ R ] 0 \small{
\ln [R]=-k t+\ln [R]_0
} ln [ R ] = − k t + ln [ R ] 0
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Derivation of integrated first rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Derivation of integrated first rate law(I/III)
Rearranging this equation
ln [ R ] [ R ] 0 = − k t or k = 1 t ln [ R ] 0 [ R ]
\begin{aligned}
& \ln \frac{[R]}{[R]_0}=-k t \\
& \text { or } k=\frac{1}{t} \ln \frac{[R]_0}{[R]}
\end{aligned}
ln [ R ] 0 [ R ] = − k t or k = t 1 ln [ R ] [ R ] 0
इस समीकरण को पुन: व्यवस्थित करना
ln [ R ] [ R ] 0 = − k t या k = 1 t ln [ R ] 0 [ R ] \small{
\begin{aligned}
& \ln \frac{[R]}{[R]_0}=-k t \\\
& \text{ या } k=\frac{1}{t} \ln \frac{[R]_0}{[R]}
\end{aligned}
} ln [ R ] 0 [ R ] = − k t या k = t 1 ln [ R ] [ R ] 0
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Derivation of integrated first rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Derivation of integrated first rate law(II/III)
At time t 1 t_1 t 1
∗ ln [ R ] 1 = − k t 1 + ∗ ln [ R ] 0 \small{
{ }^* \ln [R]_1=-k t_1+{ }^* \ln [R]_0
} ∗ ln [ R ] 1 = − k t 1 + ∗ ln [ R ] 0
At time t 2 \small{t_2} t 2
ln [ R ] 2 = − k t 2 + ln [ R ] 0 \small{
\ln [R]_2=-k t_2+\ln [R]_0
} ln [ R ] 2 = − k t 2 + ln [ R ] 0
समय t 1 t_1 t 1
∗ ln [ R ] 1 = − k t 1 + ∗ ln [ R ] 0 \small{
{ }^* \ln [R]_1=-k t_1+{ }^* \ln [R]_0
} ∗ ln [ R ] 1 = − k t 1 + ∗ ln [ R ] 0
समय t 2 \small{t_2} t 2
ln [ R ] 2 = − k t 2 + ln [ R ] 0 \small{
\ln [R]_2=-k t_2+\ln [R]_0
} ln [ R ] 2 = − k t 2 + ln [ R ] 0
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Derivation of integrated first rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Derivation of integrated first rate law(II/III)
where, [ R ] 1 \small{[R]_1} [ R ] 1 [ R ] 2 \small{[R]_2} [ R ] 2 t 1 \small{t_1} t 1 t 2 t_2 t 2
ln [ R ] 1 − ln [ R ] 2 = − k t 1 − ( − k t 2 ) ln [ R ] 1 [ R ] 2 = k ( t 2 − t 1 ) k = 1 ( t 2 − t 1 ) ln [ R ] 1 [ R ] 2 \small{
\begin{aligned}
& \ln [R]_1-\ln [R]_2=-k t_1-(-k t_2) \\\
& \ln \frac{[R]_1}{[R]_2}=k(t_2-t_1) \\\
& k=\frac{1}{(t_2-t_1)} \ln \frac{[R]_1}{[R]_2}
\end{aligned}
} ln [ R ] 1 − ln [ R ] 2 = − k t 1 − ( − k t 2 ) ln [ R ] 2 [ R ] 1 = k ( t 2 − t 1 ) k = ( t 2 − t 1 ) 1 ln [ R ] 2 [ R ] 1
ln [ R ] [ R ] 0 = − k t \small{
\ln \frac{[R]}{[R]_0}=-k t
} ln [ R ] 0 [ R ] = − k t
जहाँ, [ R ] 1 \small{[R]_1} [ R ] 1 [ R ] 2 \small{[R]_2} [ R ] 2 t 1 \small{t_1} t 1 t 2 t_2 t 2
ln [ R ] 1 − ln [ R ] 2 = − k t 1 − ( − k t 2 ) ln [ R ] 1 [ R ] 2 = k ( t 2 − t 1 ) k = 1 ( t 2 − t 1 ) ln [ R ] 1 [ R ] 2 \small{
\begin{aligned}
& \ln [R]_1-\ln [R]_2=-k t_1-(-k t_2) \\\
& \ln \frac{[R]_1}{[R]_2}=k(t_2-t_1) \\\
& k=\frac{1}{(t_2-t_1)} \ln \frac{[R]_1}{[R]_2}
\end{aligned}
} ln [ R ] 1 − ln [ R ] 2 = − k t 1 − ( − k t 2 ) ln [ R ] 2 [ R ] 1 = k ( t 2 − t 1 ) k = ( t 2 − t 1 ) 1 ln [ R ] 2 [ R ] 1
ln [ R ] [ R ] 0 = − k t \small{
\ln \frac{[R]}{[R]_0}=-k t
} ln [ R ] 0 [ R ] = − k t
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Derivation of integrated first rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Derivation of integrated first rate law(III/III)
Taking antilog of both sides
[ R ] = [ R ] 0 e − k t \small{
[R]=[R]_0 e^{-k t}
} [ R ] = [ R ] 0 e − k t
The first order rate equation can also be written in the form
k = 2.303 t log [ R ] 0 [ R ] ∗ log [ R ] 0 [ R ] = k t 2.303 \small{
\begin{aligned}
& k=\frac{2.303}{t} \log \frac{[R]_0}{[R]} \\\
& { }^* \log \frac{[R]_0}{[R]}=\frac{k t}{2.303}
\end{aligned}
} k = t 2.303 log [ R ] [ R ] 0 ∗ log [ R ] [ R ] 0 = 2.303 k t
दोनों पक्षों का एंटीलॉग लेते हैं
[ R ] = [ R ] 0 e − k t \small{
[R]=[R]_0 e^{-k t}
} [ R ] = [ R ] 0 e − k t
प्रथम क्रम दर समीकरण को भी इस रूप में लिखा जा सकता है
k = 2.303 t log [ R ] 0 [ R ] ∗ log [ R ] 0 [ R ] = k t 2.303 \small{
\begin{aligned}
& k=\frac{2.303}{t} \log \frac{[R]_0}{[R]} \\\
& { }^* \log \frac{[R]_0}{[R]}=\frac{k t}{2.303}
\end{aligned}
} k = t 2.303 log [ R ] [ R ] 0 ∗ log [ R ] [ R ] 0 = 2.303 k t
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Derivation of integrated first rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Alternative to find integrated rate law
∫ A 0 A − d [ A ] [ A ] \int _{A _0} ^{A} \frac {-d[A]} {[A]} ∫ A 0 A [ A ] − d [ A ]
= k ∫ 0 t d t ⇒ − ln [ A ] A 0 A = k ( t ) 0 t =k \int _0 ^t d t \Rightarrow - \ln [A] _ {A _0} ^{A} =k(t) _0^t = k ∫ 0 t d t ⇒ − ln [ A ] A 0 A = k ( t ) 0 t
⇒ − ln [ A ] + ( I n [ A 0 ] ) \Rightarrow - \ln [A]+(In[A _0]) ⇒ − ln [ A ] + ( I n [ A 0 ])
= k ( t − 0 ) ⇒ I n [ A 0 ] − ln [ A ] = k t ln ( [ A 0 ] [ A ] ) = k t =k(t-0) \Rightarrow In[A _0] - \ln [A]=kt \ln (\frac{[A _0]}{[A]})=k t = k ( t − 0 ) ⇒ I n [ A 0 ] − ln [ A ] = k t ln ( [ A ] [ A 0 ] ) = k t
∫ A 0 A − d [ A ] [ A ] \int _{A _0} ^{A} \frac {-d[A]} {[A]} ∫ A 0 A [ A ] − d [ A ]
= k ∫ 0 t d t ⇒ − ln [ A ] A 0 A = k ( t ) 0 t =k \int _0 ^t d t \Rightarrow - \ln [A] _ {A _0} ^{A} =k(t) _0^t = k ∫ 0 t d t ⇒ − ln [ A ] A 0 A = k ( t ) 0 t
⇒ − ln [ A ] + ( I n [ A 0 ] ) \Rightarrow - \ln [A]+(In[A _0]) ⇒ − ln [ A ] + ( I n [ A 0 ])
= k ( t − 0 ) ⇒ I n [ A 0 ] − ln [ A ] = k t ln ( [ A 0 ] [ A ] ) = k t =k(t-0) \Rightarrow In[A _0] - \ln [A]=kt \ln (\frac{[A _0]}{[A]})=k t = k ( t − 0 ) ⇒ I n [ A 0 ] − ln [ A ] = k t ln ( [ A ] [ A 0 ] ) = k t
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Alternative to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr →
Chemical Kienetics
First order graph(I/II)
For first order reaction:
[ A ] = [ A ] 0 e − k t \small{
[A]=[A]_0 e^{-k t}
} [ A ] = [ A ] 0 e − k t
Comparing it with the equation:
y = e − x \small{y = e^{-x}} y = e − x
we get,
x-axis = t \small{\text{x-axis} = t} x-axis = t
y-axis = concentration at time t \small{\text{y-axis} = \text{concentration at time t}} y-axis = concentration at time t
A first order reaction never goes to 100% completion
[ A ] = [ A ] 0 e − k t \small{
[A]=[A]_0 e^{-k t}
} [ A ] = [ A ] 0 e − k t
इसे समीकरण के साथ तुलना करते हैं:
y = e − x \small{y = e^{-x}} y = e − x
हमें मिलता है,
x-अक्ष = t \small{\text{x-अक्ष} = t} x- अक्ष = t
y-अक्ष = समय t पर संचय \small{\text{y-अक्ष} = \text{समय t पर संचय}} y- अक्ष = समय t पर संचय
पहले क्रम की प्रतिक्रिया कभी 100% पूर्णता तक नहीं जाती
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → First order graph → \rarr → → \rarr → → \rarr →
Chemical Kienetics
First order graph(II/II)
For first order reaction:
l n [ A ] t = − k t + l n [ A ] o \small{
ln[A]_t = -kt + ln[A]_o
} l n [ A ] t = − k t + l n [ A ] o
Comparing it with the equation:
y = m x + c \small{y = mx +c} y = m x + c
we get,
x-axis = t \small{\text{x-axis} = t} x-axis = t
y-axis = Natural log concentration at time t \small{\text{y-axis} = \text{Natural log concentration at time t}} y-axis = Natural log concentration at time t
s l o p e = k \small{slope = k} s l o p e = k
I n t e r c e p t = Natural log of initial concentration \small{Intercept = \text{Natural log of initial concentration}} I n t erce pt = Natural log of initial concentration
पहले क्रम की प्रतिक्रिया के लिए:
l n [ A ] t = − k t + l n [ A ] o \small{
ln[A]_t = -kt + ln[A]_o
} l n [ A ] t = − k t + l n [ A ] o
इसे समीकरण के साथ तुलना करते हैं:
y = m x + c \small{y = mx +c} y = m x + c
हमें मिलता है,
x-अक्ष = t \small{\text{x-अक्ष} = t} x- अक्ष = t
y-अक्ष = समय t पर प्राकृतिक लॉग संचयन \small{\text{y-अक्ष} = \text{समय t पर प्राकृतिक लॉग संचयन}} y- अक्ष = समय t पर प्राकृतिक लॉग संचयन
ढाल = k \small{ढाल = k} ढाल = k
अंतःकटि = प्रारंभिक संचयन का प्राकृतिक लॉग \small{अंतःकटि = \text{प्रारंभिक संचयन का प्राकृतिक लॉग}} अंतःकटि = प्रारंभिक संचयन का प्राकृतिक लॉग
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → First order graph → \rarr → → \rarr → → \rarr →
Chemical Kienetics
Gas phase first order reaction
A ( g ) A(g) A ( g ) ⟶ \longrightarrow ⟶ B ( g ) + \mathrm{B}(\mathrm{g})+ B ( g ) + C ( g ) C(g) C ( g )
1 mole of A A A
1 mole of B B B
1 mole of C C C
At t − 0 t-0 t − 0
P 0 P_0 P 0 0
0
At time
0 − P ) a t m 0-P)_{\mathrm{atm}} 0 − P ) atm P a t m \mathrm{P} _{\mathrm{atm}} P atm P a t m P _{\mathrm{atm}} P atm
k = 2.303 t log P 0 P A
k=\frac{2.303}{t} \log \frac{P_0}{P_A}
k = t 2.303 log P A P 0
ए ( जी ) ए(जी) ए ( जी ) ⟶ \longrightarrow ⟶ B ( g ) + \mathrm{B}(\mathrm{g})+ B ( g ) + सी ( जी ) सी(जी) सी ( जी )
A A A B B B C C C
t − 0 t-0 t − 0 P 0 P_0 P 0 0
0
समय पर
0 − P ) a t m 0-P)_{\mathrm{atm}} 0 − P ) atm P a t m \mathrm{P} _{\mathrm{atm}} P atm P a t m P _{\mathrm{atm}} P atm
k = 2.303 t log P 0 P A
k=\frac{2.303}{t} \log \frac{P_0}{P_A}
k = t 2.303 log P A P 0
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Gas phase first order reaction → \rarr → → \rarr →
Chemical Kienetics
Gas phase first order reaction
P A P_A P A
Total pressure ( P t ) = P \left(P _t\right)=P ( P t ) = P ( P A ) \left(P _A\right) ( P A )
Partial pressure of gaseous products ( P B + P C ) \left(P_B+P_C\right) ( P B + P C )
P t = ( P 0 − P ) + P + P P t = P 0 + P P = P t − P 0 P A = P 0 − P ∴ P A = P 0 − ( P t − P 0 ) P A = 2 P 0 − P t
\begin{aligned}
P _t & =\left(P _0-P\right)+P+P \\
& P t=P_0+P \\
P & =P_t-P_0 \\
& P_A=P_0-P \\
\therefore \quad & P _A = P _0-\left(P _t-P _0\right) \\
& P _A=2 P _0-P _t
\end{aligned}
P t P ∴ = ( P 0 − P ) + P + P Pt = P 0 + P = P t − P 0 P A = P 0 − P P A = P 0 − ( P t − P 0 ) P A = 2 P 0 − P t
P A P_A P A
कुल दबाव ( P t ) = P \left(P _t\right)=P ( P t ) = P ( P A ) \left(P _A\right) ( P A )
गैसीय उत्पादों का आंशिक दबाव ( P B + P C ) \left(P_B+P_C\right) ( P B + P C )
P t = ( P 0 − P ) + P + P P t = P 0 + P P = P t − P 0 P A = P 0 − P ∴ P A = P 0 − ( P t − P 0 ) P A = 2 P 0 − P t
\begin{aligned}
P _t & =\left(P _0-P\right)+P+P \\
& P t=P_0+P \\
P & =P_t-P_0 \\
& P_A=P_0-P \\
\therefore \quad & P _A = P _0-\left(P _t-P _0\right) \\
& P _A=2 P _0-P _t
\end{aligned}
P t P ∴ = ( P 0 − P ) + P + P Pt = P 0 + P = P t − P 0 P A = P 0 − P P A = P 0 − ( P t − P 0 ) P A = 2 P 0 − P t
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Gas phase first order reaction → \rarr → → \rarr →
Chemical Kienetics
Examples of first order reaction
Hydrogenation of ethene is an example of first order reaction.
C 2 H 4 ( g ) + H 2 ( g ) → C 2 H 6 ( g ) Rate = k [ C 2 H 4 ] \small{
\begin{aligned}
& C_2 H_4(g)+H_2(g) \to C_2 H_6(g) \\\
& \text{ Rate }=k[C_2 H_4]
\end{aligned}
} C 2 H 4 ( g ) + H 2 ( g ) → C 2 H 6 ( g ) Rate = k [ C 2 H 4 ]
All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.
88 R a 226 → 2 4 H e + 86 R n 222 Rate = k [ R a ] \small{
\begin{aligned}
& _{88}Ra^{226} \to _2^4 He+ _{86}Rn^{222} \\\
& \text{ Rate }=k[Ra]
\end{aligned}
} 88 R a 226 → 2 4 He + 86 R n 222 Rate = k [ R a ]
Decomposition of N 2 O 5 \small{N_2 O_5} N 2 O 5 N 2 O \small{N_2 O} N 2 O
इथीन का हाइड्रोजनीकरण प्रथम क्रम की प्रतिक्रिया का उदाहरण है।
C 2 H 4 ( g ) + H 2 ( g ) → C 2 H 6 ( g ) दर = k [ C 2 H 4 ] \small{
\begin{aligned}
& C_2 H_4(g)+H_2(g) \to C_2 H_6(g) \\\
& \text{ दर }=k[C_2 H_4]
\end{aligned}
} C 2 H 4 ( g ) + H 2 ( g ) → C 2 H 6 ( g ) दर = k [ C 2 H 4 ]
सभी प्राकृतिक और कृत्रिम रेडियोधर्मी अस्थिर नाभिकों का क्षय प्रथम क्रम के गतिकी द्वारा होता है।
88 R a 226 → 2 4 H e + 86 R a 222 दर = k [ R a ] \small{
\begin{aligned}
& _{88}Ra^{226} \to _2^4 He+ _{86}Ra^{222} \\\
& \text{ दर }=k[Ra]
\end{aligned}
} 88 R a 226 → 2 4 He + 86 R a 222 दर = k [ R a ]
N 2 O 5 \small{N_2 O_5} N 2 O 5 N 2 O \small{N_2 O} N 2 O
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Examples of first order reaction → \rarr →
Chemical Kinetics
Example
Question : The initial concentration of N 2 O 5 \mathrm{N} _2 \mathrm{O} _5 N 2 O 5 N 2 O 5 ( g ) → 2 N O 2 ( g ) + 1 / 2 O 2 \mathrm{N} _2 \mathrm{O} _5(\mathrm{g}) \rightarrow 2 \mathrm{NO} _2(\mathrm{g})+1 / 2 \mathrm{O} _2 N 2 O 5 ( g ) → 2 NO 2 ( g ) + 1/2 O 2 1.24 × 10 − 2 m o l L − 1 1.24 \times 10^{-2} \mathrm{mol} \mathrm{L}^{-1} 1.24 × 1 0 − 2 mol L − 1 318 K 318 \mathrm{K} 318 K N 2 O 5 \mathrm{N} _2 \mathrm{O} _5 N 2 O 5 0.20 × 10 − 2 m o l L − 1 0.20 \times 10^{-2} \mathrm{mol} \mathrm{L}^{-1} 0.20 × 1 0 − 2 mol L − 1 318 K 318 \mathrm{K} 318 K
सवाल : निम्नलिखित प्रथम क्रम प्रतिक्रिया में N 2 O 5 \mathrm{N} _2 \mathrm{O} _5 N 2 O 5 N 2 O 5 ( g ) → 2 N O 2 ( g ) + 1 / 2 O 2 \mathrm{N} _2 \mathrm{O} _5(\mathrm{g}) \rightarrow 2 \mathrm{NO} _2(\mathrm{g})+1/2 \mathrm{O} _2 N 2 O 5 ( g ) → 2 NO 2 ( g ) + 1/2 O 2 1.24 × 10 − 2 m o l L थ ा − 1 1.24\times 10^{-2}\mathrm{mol}\mathrm{L} था ^{-1} 1.24 × 1 0 − 2 mol L थ ा − 1 318 K 318 \mathrm{K} 318 K N 2 O 5 \mathrm{N} _2\mathrm{O} _5 N 2 O 5 0.20 × 10 − 2 m o l L − 1 0.20\times 10^{-2}\mathrm{mol}\mathrm{L}^{-1} 0.20 × 1 0 − 2 mol L − 1 318 K 318 \mathrm{K} 318 K
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Example
Chemical Kinetics
Example
Solution : For a first order reaction
log [ R ] 1 [ R ] 2 = k t 2 − t 1 2.303 k = 2.303 t 2 − t 1 log [ R ] 1 [ R ] 2 = 2.303 ( 60 m i n − 0 m i n ) log 1.24 × 10 − 2 m o l L − 1 0.20 × 10 − 2 m o l L − 1 = 2.303 60 log 6.2 m i n − 1 k = 0.0304 m i n − 1
\begin{aligned}
\log \frac{[\mathrm{R}]_1}{[\mathrm{R}]_2} & =\frac{kt_2-t_1}{2.303} \\
k & =\frac{2.303}{t_2-t_1} \log \frac{[R]_1}{[R]_2} \\
& =\frac{2.303}{(60 \mathrm{min}-0 \mathrm{min})} \\
& \log \frac{1.24 \times 10^{-2} \mathrm{mol} \mathrm{L}^{-1}}{0.20 \times 10^{-2} \mathrm{mol} \mathrm{L}^{-1}} \\
& =\frac{2.303}{60} \log 6.2 \mathrm{min}^{-1} \\
k & =0.0304 \mathrm{min}^{-1}
\end{aligned}
log [ R ] 2 [ R ] 1 k k = 2.303 k t 2 − t 1 = t 2 − t 1 2.303 log [ R ] 2 [ R ] 1 = ( 60 min − 0 min ) 2.303 log 0.20 × 1 0 − 2 mol L − 1 1.24 × 1 0 − 2 mol L − 1 = 60 2.303 log 6.2 min − 1 = 0.0304 min − 1
समाधान: प्रथम कोटि की प्रतिक्रिया के लिए
log [ R ] 1 [ R ] 2 = k t 2 − t 1 2.303 k = 2.303 t 2 − t 1 log [ R ] 1 [ R ] 2 = 2.303 ( 60 m i n − 0 m i n ) log 1.24 × 10 − 2 m o l L − 1 0.20 × 10 − 2 m o l L − 1 = 2.303 60 log 6.2 m i n − 1 k और = 0.0304 m i n − 1
\begin{aligned}
\log \frac{[\mathrm{R}]_1}{[\mathrm{R}]_2} & =\frac{kt_2-t_1}{2.303} \\
k & =\frac{2.303}{t_2-t_1} \log \frac{[R]_1}{[R]_2} \\
& =\frac{2.303}{(60 \mathrm{min}-0 \mathrm{min})} \\
& \log \frac{1.24 \times 10^{-2} \mathrm{mol} \mathrm{ L}^{-1}}{0.20 \times 10^{-2} \mathrm{mol} \mathrm{L}^{-1}} \\
& =\frac{2.303}{60} \log 6.2 \mathrm{min}^{-1} \\
k और =0.0304 \mathrm{min}^{-1}
\end{aligned}
log [ R ] 2 [ R ] 1 k k और = 0.0304 min − 1 = 2.303 k t 2 − t 1 = t 2 − t 1 2.303 log [ R ] 2 [ R ] 1 = ( 60 min − 0 min ) 2.303 log 0.20 × 1 0 − 2 mol L − 1 1.24 × 1 0 − 2 mol L − 1 = 60 2.303 log 6.2 min − 1
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Example
Chemical Kinetics
Example
Question : The following data were obtained during the first order thermal decomposition of N 2 O 5 ( g ) \mathrm{N} _2 \mathrm{O} _5(\mathrm{g}) N 2 O 5 ( g )
S.No.
Time/s
Total Pressure/(atm)
1.
0
0.5
2.
100
0.512
Calculate the rate constant.
प्रश्न: स्थिर आयतन पर N 2 O 5 ( g ) \mathrm{N} _2 \mathrm{O} _5(\mathrm{g}) N 2 O 5 ( g )
S.No.
Time/s
Total Pressure/(atm)
1.
0
0.5
2.
100
0.512
दर स्थिरांक की गणना करें.
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Example
Chemical Kinetics
Example
Solution : Let the pressure of N 2 O 5 ( g ) \mathrm{N} _2 \mathrm{O} _5(\mathrm{g}) N 2 O 5 ( g ) 2 x 2 \mathrm{x} 2 x N 2 O 5 \mathrm{N} _2 \mathrm{O} _5 N 2 O 5 N 2 O 4 ( g ) \mathrm{N} _2 \mathrm{O} _4(\mathrm{g}) N 2 O 4 ( g ) O 2 ( g ) \mathrm{O} _2(\mathrm{g}) O 2 ( g ) N 2 O 4 ( g ) \mathrm{N} _2 \mathrm{O} _4(\mathrm{g}) N 2 O 4 ( g ) 2 x 2 \mathrm{x} 2 x O 2 ( g ) \mathrm{O} _2(\mathrm{g}) O 2 ( g ) x \mathrm{x} x
2 N 2 O 5 g 2 \mathrm{N} _2 \mathrm{O} _5 \mathrm{g} 2 N 2 O 5 g 2 N 2 O 4 g 2 \mathrm{N} _2 \mathrm{O} _4 \mathrm{g} 2 N 2 O 4 g O 2 g \mathrm{O} _2 \mathrm{g} O 2 g
Start t = 0 t=0 t = 0
0.5 a t m 0.5 \mathrm{atm} 0.5 atm 0 a t m 0 \mathrm{atm} 0 atm 0 a t m 0 \mathrm{atm} 0 atm
At time t t t
( 0.5 − 2 x ) a t m (0.5-2 \mathrm{x}) \mathrm{atm} ( 0.5 − 2 x ) atm 2 x 2 \mathrm{x} 2 x x a t m \mathrm{x} \mathrm{atm} x atm
समाधान: मान लीजिए कि N 2 O 5 ( g ) \mathrm{N} _2 \mathrm{O} _5(\mathrm{g}) N 2 O 5 ( g ) 2 x 2 \mathrm{x} 2 x N 2 O 5 \mathrm{N} _2 \mathrm{O} _5 N 2 O 5 N 2 O 4 ( g ) \mathrm{N} _2 \mathrm{O} _4(\mathrm{g}) N 2 O 4 ( g ) O 2 ( g ) \mathrm{O} _2(\mathrm{g}) O 2 ( g ) N 2 O 4 ( g ) \mathrm{N} _2 \mathrm{O} _4(\mathrm{g}) N 2 O 4 ( g ) 2 x सेबढ़जाताहै 2 \mathrm{x} से बढ़ जाता है 2 x सेबढ़जाताहै O 2 ( g ) \mathrm{O} _2(\mathrm{g}) O 2 ( g ) x \mathrm{x} x
2 N 2 O 5 g 2 \mathrm{N} _2 \mathrm{O} _5 \mathrm{g} 2 N 2 O 5 g 2 N 2 O 4 g 2 \mathrm{N} _2 \mathrm{O} _4 \mathrm{g} 2 N 2 O 4 g O 2 g \mathrm{O} _2 \mathrm{g} O 2 g
Start t = 0 t=0 t = 0
0.5 a t m 0.5 \mathrm{atm} 0.5 atm 0 a t m 0 \mathrm{atm} 0 atm 0 a t m 0 \mathrm{atm} 0 atm
At time t t t
( 0.5 − 2 x ) a t m (0.5-2 \mathrm{x}) \mathrm{atm} ( 0.5 − 2 x ) atm 2 x 2 \mathrm{x} 2 x x a t m \mathrm{x} \mathrm{atm} x atm
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Example
Chemical Kinetics
Example
p t = p N 2 O 5 + p N 2 O 4 + p O 2 p_{t} = p_{N_{2} O_{5} } + p_ {N_{2} O_{4} } + p_{O_{2}} p t = p N 2 O 5 + p N 2 O 4 + p O 2
= ( 0.5 − 2 x ) + 2 x + x = 0.5 + x =(0.5-2 \mathrm{x})+2 \mathrm{x}+\mathrm{x}=0.5+\mathrm{x} = ( 0.5 − 2 x ) + 2 x + x = 0.5 + x
x p t 0.5 \mathrm{x} p_t \quad 0.5 x p t 0.5
p N 2 O 5 = 0.5 − 2 x p _{\mathrm{N}_2 \mathrm{O}_5}=0.5-2 \mathrm{x} p N 2 O 5 = 0.5 − 2 x
= 0.5 − 2 ( p t − 0.5 ) = 1.5 − 2 p t =0.5-2\left(p _{\mathrm{t}}-0.5\right)=1.5-2 p_t = 0.5 − 2 ( p t − 0.5 ) = 1.5 − 2 p t
At t = 100 s ; p t = 0.512 a t m \text { At } t=100 \mathrm{s} ; p_{\mathrm{t}}=0.512 \mathrm{atm} At t = 100 s ; p t = 0.512 atm
p t = p N 2 O 5 + p N 2 O 4 + p O 2 p_{t} = p_{N_{2} O_{5} } + p_ {N_{2} O_{4} } + p_{O_{2}} p t = p N 2 O 5 + p N 2 O 4 + p O 2
= ( 0.5 − 2 x ) + 2 x + x = 0.5 + x =(0.5-2 \mathrm{x})+2 \mathrm{x}+\mathrm{x}=0.5+\mathrm{x} = ( 0.5 − 2 x ) + 2 x + x = 0.5 + x
x p t 0.5 \mathrm{x} p_t \quad 0.5 x p t 0.5
p N 2 O 5 = 0.5 − 2 x p _{\mathrm{N}_2 \mathrm{O}_5}=0.5-2 \mathrm{x} p N 2 O 5 = 0.5 − 2 x
= 0.5 − 2 ( p t − 0.5 ) = 1.5 − 2 p t =0.5-2\left(p _{\mathrm{t}}-0.5\right)=1.5-2 p_t = 0.5 − 2 ( p t − 0.5 ) = 1.5 − 2 p t
At t = 100 s ; p t = 0.512 a t m \text { At } t=100 \mathrm{s} ; p_{\mathrm{t}}=0.512 \mathrm{atm} At t = 100 s ; p t = 0.512 atm
Algorithm to find integrated rate law → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → → \rarr → Example
Chemical Kienetics Topics to be covered Algorithm to find integrated rate law Zero Order Reaction Integrated rate law derivation Alternative to find integrated rate law Zero order graph Example of Zero order reaction First order reaction Derivation of integrated first rate law Alternative to find integrated rate law First order graph Gas phase first order reaction Examples of first order reaction एकीकृत दर कानून खोजने के लिए एल्गोरिदम शून्य आदेश प्रतिक्रिया एकीकृत दर कानून व्युत्पत्ति एकीकृत दर कानून खोजने का विकल्प शून्य क्रम ग्राफ शून्य कोटि अभिक्रिया का उदाहरण प्रथम कोटि की प्रतिक्रिया एकीकृत प्रथम श्रेणी कानून की व्युत्पत्ति एकीकृत दर कानून खोजने का विकल्प प्रथम क्रम ग्राफ गैस चरण प्रथम क्रम प्रतिक्रिया प्रथम कोटि अभिक्रिया के उदाहरण Algorithm to find integrated rate law $\rarr$ Zero Order Reaction $\rarr$ Integrated rate law derivation $\rarr$ Alternative to find integrated rate law $\rarr$ Zero order graph $\rarr$ Example of Zero order reaction $\rarr$ First order reaction $\rarr$ Derivation of integrated first rate law $\rarr$ Alternative to find integrated rate law $\rarr$ First order graph $\rarr$ Gas phase first order reaction $\rarr$ Examples of first order reaction $\rarr$ Example
