Chemical Kinetics - class-XII
Covers reaction rates, rate expressions, reaction order, rate constants, and graphical determination methods for chemical reactions
Questions
When a chemical reaction takes place, during the course of the reaction the rate of reaction?
- Keeps on increasing with time
- Remains constant with time
- Keeps on decreasing with time
- Shows irregular trend with time
In a reaction $2HI \rightarrow H _{2} + I _{2}$, the concentration of $HI$ decreases from $0.5\ mol\ L^{-1}$ to $0.4\ mol\ L^{-1}$ in $10$ minutes. What is the rate of reaction during this interval?
- $5\times 10^{-3} M\ min^{-1}$
- $2.5\times 10^{-3} M\ min^{-1}$
- $5\times 10^{-2} M\ min^{-1}$
- $2.5\times 10^{-2} M\ min^{-1}$
Which of the following statements is correct?
- The rate of a reaction decreases with passage of time as the concentration of reactants decreases.
- The rate of a reaction is same at any time during the reaction.
- The rate of a reaction is independent of temperature change.
- The rate of a reaction decreases with increase in concentration of reaction(s).
The rate constant of a zero order reaction is 0.2 mol $d{ m }^{ -3 }{ h }^{ -3 }$. If the concentration minutes is 0.05 mol $d{ m }^{ -3 }$. Then its initial concentration would be:
- $6.05 mol d{ m }^{ -3 }$
- $0.15 mol d{ m }^{ -3 }$
- $0.25 mol d{ m }^{ -3 }$
- none of the above
Consider the chemical reaction:
$N _2(g)+3H _2(g)\rightarrow 2NH _3(g)$
The rate of this reaction can be expressed; in terms of time and of concentration of $N _2(g), H _2(g)$ $NH _3(g)$. Identify the correct relationship amongst the rate expressions.
- Rate $=-\dfrac{d[N _2]}{dt}=-\dfrac{1}{3}\dfrac{d[H _2]}{dt}=+\dfrac{1}{2}\dfrac{d[NH _3]}{dt}$
- Rate $=-\dfrac{d[N _2]}{dt}=-\dfrac{3d[H _2]}{dt}=\dfrac{2d[NH _3]}{dt}$
- Rate $=-\dfrac{d[N _2]}{dt}=-\dfrac{1}{3}\dfrac{d[H _2]}{dt}=\dfrac{d[NH _3]}{dt}$
- Rate $=-\dfrac{d[N _2]}{dt}=\dfrac{d[H _2]}{dt}=\dfrac{d[NH _3]}{dt}$
Rate of formation of $SO _3$ in the following reaction $2SO _2+O _2\rightarrow 2SO _3$ is $100g$ $min^{-1}$.
- $50g$ $min^{-1}$
- $40g$ $min^{-1}$
- $200g$ $min^{-1}$
- $20g$ $min^{-1}$
If concentration of reactants is increased by a factor x then the rate constant k becomes:
- $\ln{\frac{k}{x}}$
- $\frac{k}{x}$
- $k+x$
- $k$
An aqueous solution of $CH _3COOH$ has a pH = $3$ and acid dissociation constant of $CH _3COOH$ is $10^{-5}$. What will be the concentration of acid taken initially?
- $0.1$M
- $0.11$M
- $0.09$M
- $0.101$M
For the non-equilibrium process, $A + B \rightarrow Products$, the rate is first order with respect to $A$ and second-order with respect to $B$. If $1.0$ mole each of $A$ and $B$ are introduced into a 1-litre vessel and the initial rate was $1.0 \times 10^{-2}$ mol/litre-sec. The rate (in mol $litre^{-1} sec^{-1}$) when half of the reactants have been used:
- $1.2 \times 10^{-3}$
- $1.2 \times 10^{-2}$
- $2.5 \times 10^{-4}$
- none of these
| Time | 0 | 5min | 10min | 15min |
|---|---|---|---|---|
| [A] | 20mol | 18mol | 16mol | 16 mol |
For the reaction $A\longrightarrow Products$; $\frac { -d[A] }{ dt } =k$ and at different time interval, IAI values are given. At $20$ minute, rate will be :
- $12 mol /min$
- $10 mol/min$
- $8 mol/min$
- $0.4 mol/min$
$H _2 + l _2 \rightarrow 2 Hl$ (An elementary reaction)
If the volume of the container containing the gaseous mixture is increased to two times, then final rate of the reaction
- Become four time
- Become $\dfrac{1}{4} th$ of the original rate
- Become $2$ times
- Become $\dfrac{1}{2}$ of the original rate
Assuming an element reaction $H _2O _2+ 3I^-+ 2H^+\to 2H _2O+ I _3^-.$ The effect on the rate of this reaction brought about by doubling the concentration of $I^-$ without changing the order?
- The rate would increases by a factor of $3$
- The rate would increase by a factor of $8$
- The rate would decrease by a factor of $1/3$
- The rate would increase by a factor of $9$
Instanteneous rate of reaction can be found be :
- slope of a rate of reaction vs time
- slope of a concentration vs time graph
- taking any two points on the graph
- both $B$ and $C$
For the reaction, $2{ N } _{ 2 }{ O } _{ 5 }\left( g \right) \longrightarrow 4N{ O } _{ 2 }\left( g \right) +{ O } _{ 2 }\left( g \right) $, if the concentration of $N{ O } _{ 2 }$ increases by $5.2\times { 10 }^{ -3 }M$ in $100$ sec, then the rate of reaction is:
- $1.3\times { 10 }^{ -5 }M{ s }^{ -1 }$
- $5\times { 10 }^{ -4 }M{ s }^{ -1 }$
- $7.6\times { 10 }^{ -4 }M{ s }^{ -1 }$
- $2\times { 10 }^{ -3 }M{ s }^{ -1 }$
- $2.5\times { 10 }^{ -5 }M{ s }^{ -1 }$
A reaction is represented as $2A + B \mapsto 2C + 3D$. The concentration of C at 10 s is 4 moles $l^{-1}$. The concentration of C at 20 seconds is 5.2 moles $l^{-1}$. The rate of reaction of B in the same time interval could be :
- $-0.12$ mole $l^{-1} S^{-1}$
- $-0.6$ mole $l^{-1} S^{-1}$
- $-0.06$ mole $l^{-1} S^{-1}$
- $-1.2$ mole $l^{-1} S^{-1}$
For $S{O _2}C{l _{2\left( g \right)}} \to S{O _{2\left( g \right)}} + C{l _{2\left( g \right)}},$ Pressures of $S{O _2}C{l _2}$ at $t = 0$ and $t = 20$ minutes respectively are $700mm$ and $350mm.$ When $\log \left( {{P _0}/p} \right)$ is plotted against time ($t$), slope equals to:
- $1.505 \times {10^{ - 2}}{s^{ - 1}}$
- $1.202 \times {10^{ - 3}}{\min ^{ - 1}}$
- $1.505 \times {10^{ - 2}}{\min ^{ - 1}}$
- $0.3465\ {\min ^{ - 1}}$
From the concentrations of R at different times given below. Determine the average rate of the reaction range: R $\rightarrow$ P in given intervals of time.
| t (s) | 0 | 5 | 10 | 20 | 30 |
|---|---|---|---|---|---|
| $10^{-3}, \times, [R] ,(mol, L^{-1})$ | 160 | 80 | 40 | 10 | 2.5 |
- $3.5\times\,10^{2}$ to $0.42 \, \times\, 10^{2}$ $mol.L^{-1}\, s^{1}$
- $7\times\,10^{2}$ to $0.84 \, \times\, 10^{2}$ $mol.L^{-1}\, s^{1}$
- $8\times\,10^{3}$ to $0.37 \, \times\, 10^{3}$ $mol.L^{-1}\, s^{1}$
- $16\times\,10^{3}$ to $0.75 \, \times\, 10^{3}$ $mol.L^{-1}\, s^{1}$