SATHEE: Chemical Kinetics Question 291

Chemical Kinetics Question 291

Options:

A) $k = k^{'}$

B) $2 k = k^{'}$

C) $k = 2 k^{'}$

D) $k = 4 k^{'}$

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Answer:

Correct Answer: B

Solution:

[b] Rate of disappearance of reactant= Rate of appearance of products $- \frac{1}{2} \frac{d [N_{2} O_{5}]}{d t} = \frac{1}{4} \frac{d [N O_{2}]}{d t}$

$\frac{1}{2} k [N_{2} O_{5}] = \frac{1}{4} k^{'} [N_{2} O_{5}]$

$\frac{k}{2} = \frac{k^{'}}{4}$
$∴ k^{'} = 2 k$

Chemical Kinetics Question 291

Options:

Answer:

Solution:

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Chemical Kinetics Question 291

Question: For the reaction, 2N2O5→4NO2+O2 the rate equation can be expressed in two ways −d[N2O5]dt=k[N2O5] and +d[NO2]dt=k′[N2O5] k and k’ are related as:

Options:

Answer:

Solution:

Engineering Preparation

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Important Resources

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Syllabus

Test Platform

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Question: For the reaction, $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ the rate equation can be expressed in two ways $- \frac{d [N_{2} O_{5}]}{d t} = k [N_{2} O_{5}]$ and $+ \frac{d [N O_{2}]}{d t} = k^{'} [N_{2} O_{5}]$ k and k’ are related as: