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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Chemistry Test1.
Derive integrated rate law for a zero order reaction A\(\longrightarrow \) product.
2.
Define rate law and rate constant.
3.
Define average rate and instantaneous rate.
4.
The rate of the reaction X + 2y→ product is 4 x 10-3 mol L-1S-1, if [X] = [Y] = 0.2M and rate constant at 400K is 2 x 10-2s-1, What is the overall order of the reaction.
5.
What is the order with respect to each of the reactant and overall order of the following reactions?
a) 5Br-(aq)+BrO3-(aq)+6H+(aq) ➝3Br2(l)+3H2O(l)
The experimental rate law is Rate = k [Br−][BrO3−][H+]2
b) CH3CHO(g)\(\overset { \Delta }{ \longrightarrow } \) CH4(g)+CO(g) the experimental rate law is
Rate =K[CH3CHO]\(\frac{3}{2}\)
1.
A reaction in which the rate is independent of the concentration of the reactant over a wide range of concentration is called a zero order reaction.
Let us consider the following by hypothetical zero order reaction.
A⟶ product
The rate law can be written as,
Rate = k[A]0
\(\frac { -d\left[ A \right] }{ dt } =k(1)\ \ \ \therefore \left( { \left[ A \right] }^{ 0 }=1 \right) \)
\(\Rightarrow -d\left[ A \right] =kdt\)
Integrate the above equation between the limits of [A0] at zero time and [A] at some later time 't',
\(-\int _{ \left[ { A }_{ 0 } \right] }^{ \left[ A \right] }{ d\left[ A \right] } =k\int _{ 0 }^{ t }{ dt } \)
\(-{ \left( \left[ A \right] \right) }_{ \left[ { A }_{ 0 } \right] }^{ \left[ A \right] }=k{ \left( t \right) }_{ 0 }^{ t }\)
[A0] - [A] = kt
\(k=\frac { \left[ { A }_{ 0 } \right] -\left[ A \right] }{ t } \)
Straight line equation y = mx + c
ie., \(\left[ A \right] =-kt+\left[ { A }_{ 0 } \right] \)
⇒ y = c + mx
A plot [A] vs time gives a straight line with a slope of -k and y intercept of [A0]
2.
Rate law:
(i) Rate law or rate equation is an expression which relates the rate of a reaction with rate constant and the concentration of reactants.
(ii) For xA + yB → products
(iii) The rate law is r = k[A]m[B]n
Rate constant:
(i) It is the rate of the reaction when the concentration of the reactants are taken unity.
In above rate law if[A] = [B] = 1, rate constant k = Rate
3.
Average rate of reaction:
The average rate is defined as the ratio of change in the final concentration of reactants and the initial concentration of reactants over the entire time period of reaction.
Average rate=\(\frac{-[Final \space concentration \space of \space reactants - Initial \space concentration \space of \space reactants}{(Change \space in \space time)}\)
R \(=\frac{-\left(\left[A_{2}\right]-\left[A_{1}\right]\right)}{\left(t_{2}-t_{1}\right)}=-\left(\frac{\Delta[A]}{\Delta t}\right)\)
[A]1 = Concentration of reactant A1 at time t1
[A]2 = Concentration of reactant A2 at time t2
Instantaneous rate of reaction:
The rate of reaction at any particular instant during the course of reaction is called as instantaneous rate.
Instantaneous rate \(=(\text { Average rate })_{\Delta t \rightarrow 0}\)
Rate of the reaction \(=\left(\frac{-\Delta \mathrm{A}}{\Delta \mathrm{t}}\right)\)
4.
Rate = K[X]n[y]m
4 x 10-3 mol L-1s-1= 2 x 10-2s-1(0.2mol L-1)n(0.2mol L-1)m
\(\frac { 4\times { 10 }^{ -3 }mol\quad { L }^{ -1 }{ s }^{ -1 } }{ 2\times { 10 }^{ -2 }{ s }^{ -1 } } =\) (0.2)n+m(mol L-1)n+m
0.2(mol L-1) = (0.2)n+m(mol L-1)n+m
Comparing the powers on both sides
The overall order of the reaction n + m = 1
5.
a) First order with respect to Br−, first order with respect to BrO3− and second order with respect to H+. Hence the overall order of the reaction is equal to 1 + 1 + 2 = 4
b) Order of the reaction with respect to acetaldehyde is \(\frac{3}{2}\) and overall order is also \(\frac{3}{2}\)
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