Chemical Kinetics MCQs for NEET — Chemistry Questions with Answers

(d) Given, order of reaction = 1
Rate of reaction at 10 s = 0.04 mol L^-1 s^-1
Rate of reaction at 20 s = 0.03 mol L^-1s^-1

$\therefore$Half-lite period (t_1/2) = ?
We have the equation for rate-constant 'k' in first
order reaction.

                      $$\begin{gathered} \mathrm{k}=\frac{2.303}{t_2-t_1}\log \frac{{\mathrm{A}}_{\mathrm{t}}}{{\mathrm{A}}_0} \\ =\frac{2.303}{20-10}\log \frac{0.04}{0.03} \\ = \frac{2.303}{10}\times 0.124 \\ \mathrm{k}= 0.028 {\mathrm{s}}^{-1} \\ \mathrm{now} {\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} =\frac{0.693}{\mathrm{k}}=\frac{0.693}{0.028}=24.7 \mathrm{s} \end{gathered}$$

The Arrhenius equation, k = A e⁻Eᵃ/RT, relates the rate constant (k) to temperature (T) and activation energy (Eᵃ). Plotting ln(k) vs. 1/T gives a straight line with a slope of -Eᵃ/R, allowing the determination of activation energy from the slope.

(b) For a zero order reaction t_1/2 is directly proportional to the initial concentration of the reactant [R]₀

          t_1/2 $\propto$[R]₀

For a first order reaction

 k= 2.303/t. log[R]₀/[R] at t_1/2, [R] =[R]₀/2

So, the above equation becomes

          K=2.303/t_1/2 .log[R]₀/([R]₀/2)

            t_1/2 = 2.303/K = log2 = 2.303/K x .3010

            t_1/2 = .693/K

i.e, half life period is independent of initial concentration of a reactant

Key Concept: For a zero order reaction unit of rate constant is molar per second. Hence, we can easily calculate concentration of B after 20 min by the following formula,

x = Kt

X = Kt = 0.6 x 10^-3 x 20 x 60= 0.72M

(c) Given, initial temperature,
T₁=20+273=293K
Final temperature
T₂=35+273=308K

R=8.314 J mol^-1 K^-1

Since, rate becomes double on raising temperature,
∴ r₂=2r₁ or r₂/r₁=2

As rate constant, k∝r
∴ k₂/k₁=2

From Arrnhenius equation, we know that

log k₂/k₁= -E_{a______} [T₁-T₂/T₁T₂]
                 2.303R

log 2= -E_{a______}         [293-308/293x308]
            2.303x8.314

0.3010=   -E_{a______}         [-15/293x308]
            2.303x8.314

∴ E_a=0.3010x2.303x8.314x293x308
                           15

=34673.48 J mol^-1=34.7 kJ mol^-1

(b) For 10° rise in temperature, n=1 so rate = 2ⁿ = 2¹ = 2

When temperature is increased from 10°C to 100°C, change in temperature = 100-10 = 90°C, i.e., n= 9
So, rate = 2⁹ 512 times

Alternate method With every 10° rise in temperature, rate becomes double, so
$\frac{r\text{'}}{r}=2^{\left(\frac{100 - 10}{10}\right)}$ = 2⁹ = 512 times.

Order of reaction may be zero, whole number or fraction number.

Key Idea Rate of disappearance of reactant=rate of appearance of product

$$\begin{gathered} \mathrm{or} -\frac{1}{\mathrm{stoichiometric} \mathrm{coefficient} \\ \mathrm{of} \mathrm{reactant}}\frac{\mathrm{d}\left[\mathrm{reactant}\right]}{\mathrm{dt}} \\ =+\frac{1}{\mathrm{stoichiometric} \mathrm{coefficient} \\ \mathrm{of} \mathrm{product}}\frac{\mathrm{d}\left[\mathrm{product}\right]}{\mathrm{dt}} \\ \mathrm{For} \mathrm{the} \mathrm{reaction}, \\ {\mathrm{N}}_2{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NO}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right) \\ \frac{-\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{dt}}=+\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{dt}}=+\frac{2\mathrm{d}\left[{\mathrm{O}}_2\right]}{\mathrm{dt}} \\ \therefore \frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{dt}}=-2\frac{\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{dt}} \\ =2\times 6.25\times {10}^{-3} \mathrm{mol} {\mathrm{L}}^{-1}{\mathrm{s}}^{-1} \\ =12.5\times {10}^{-3} \mathrm{mol} {\mathrm{L}}^{-1}{\mathrm{s}}^{-1} \\ =1.25\times {10}^{-2}\mathrm{mol} {\mathrm{L}}^{-1}{\mathrm{s}}^{-1} \\ \frac{\mathrm{d}\left[{\mathrm{O}}_2\right]}{\mathrm{dt}}=-\frac{\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{dt}}\times \frac{1}{2} \\ =\frac{6.25\times {10}^{-3} \mathrm{mol} {\mathrm{L}}^{-1}{\mathrm{s}}^{-1}}{2} \\ =3.125\times {10}^{-3} \mathrm{mol} {\mathrm{L}}^{-1}{\mathrm{s}}^{-1} \end{gathered}$$

For an endothermic reaction, the enthalpy change (ΔH) is positive. The activation energy (Ea) is always greater than or equal to ΔH. The minimum value of Ea is when it is equal to ΔH. Therefore, the minimum value of Ea will be more than ΔH.