Chemical Kinetics Question 337
Question: In a chemical reaction A is converted into B. The rates of reaction, starting with initial concentrations of A as $ 2\times {10^{-3}}M $ and $ 1\times {10^{-3}}M, $ are equal to $ 2.40\times {10^{-4}}M{s^{-1}} $ and $ 0.60\times {10^{-4}}M{s^{-1}} $ respectively. The order of reaction with respect to reactant A will be
Options:
A) 0
B) 1.5
C) 1
D) 2
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ a\xrightarrow{{}}B $
$ \begin{aligned} & \begin{matrix} \text{Initial concentration} & \text{Rate of reaction} \\ \end{matrix} \\ & \begin{matrix} 2\times {10^{-3}}M & ,2.40\times {10^{-4}}M{s^{-1}} \\ \end{matrix} \\ & \begin{matrix} 1\times {10^{-3}}M & 0.60\times {10^{-4}}M{s^{-1}} \\ \end{matrix} \\ \end{aligned} $ rate of reaction $ r=k{{[A]}^{x}} $ where x = order of reaction hence $ 2.40\times {10^{-4}}=k{{[ 2\times {10^{-3}} ]}^{x}} $ ……(i) $ 0.60\times {10^{-4}}=k{{[1\times {10^{-3}}]}^{x}} $ ……(ii) On dividing eqn.(i) from eqn. (ii) we get $ 4={{(2)}^{x}} $
$ \therefore ~~x=2 $ i.e. order of reaction = 2
