Chemistry MCQs for NEET — Practice Questions with Answers

A + B $\to$P

     -d[A]/dt = K[A]=1.85x10⁴x[A] ....(1)

     Assuming reaction to be of second order 

     -d[A]/dt = k[A][B] = -d[A]/dt = k'[A][0.1]....(2)

     on dividing eqn (1) by (2)

     1 = 1.85x10⁴/K'x0.1 = K' = -1.85 x 10⁵ L/mol sec

 (4) t_1/2 ∝1/a^n-1

ΔH = (Ea)_t -(Ea)_b 

54 = 57.2 - x

x = 3.2 kJ

t_99.9 = (2.303/k)log (100/100-99.9) = 6.909/k ....

     Now, t_1/2 = (2.303/k)log (100/100-50) = (2.303/k)log2 ....(2) 

     From and (2) 

     t_99.9 / t_1/2 =10

k = (2.303/40)log(0.1/0.005)

      k = (2.303/40))x1.3

    Rate = k[X] = (2.303/40))x1.3x0.01

            = 7.5 x 10^-4 M min^-1

(d) We know that, slowest step is the rate determining step.

$\therefore$ Rater(r) = K1[X2][Y2]                            ......(i)

Now, from equation. (i), I.e.

        $$\begin{gathered} {\mathrm{X}}_2 \stackrel{ }{\rightleftharpoons } 2\mathrm{X}\left(\mathrm{fast}\right) \\ {\mathrm{K}}_{\mathrm{eq}} = \frac{{\left[\mathrm{X}\right]}^2}{\left[{\mathrm{X}}_2\right]} \Rightarrow \left[\mathrm{X}\right] = {\left\{{\mathrm{K}}_{\mathrm{eq}}\left[{\mathrm{X}}_2\right]\right\}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$       .....(ii)

Now, substitute the value of[X] from equation. (ii) in equation. (i), we get

$$\begin{gathered} \mathrm{Rate}\left(\mathrm{r}\right) = {\mathrm{K}}_1\left({\mathrm{K}}_{\mathrm{eq}}{)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right[{\mathrm{X}}_2{]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[{\mathrm{Y}}_2\right] \\ = \mathrm{K}\left[{\mathrm{X}}_2{]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right[{\mathrm{Y}}_2] \\ \therefore \mathrm{Order} \mathrm{of} \mathrm{reaction} = \frac{1}{2}+1=\frac{3}{2}=1.5 \end{gathered}$$

(b) For a first order reaction,

Rate constant (k) = 2.303/t . log(a/a-x)

where, a = initial concentration

a-x = concentration  after  time 't'

t= time in 'sec'

Given, a= 20 g, a-x = 5g, k=10^-2

.^.. t =2.303/10^-2 . log(20/5) = 138.6 s

Alternatively,

Half-life for the first order reaction,

    t_1/2/2 = 0.693/k = 0.693/10^-2 = 69.3s

Two half-lives  are required for the reduction of 20 g of reactant into 5g

    20g $\stackrel{{\mathrm{t}}_{1/2}}{\to }$10g$\stackrel{{\mathrm{t}}_{1/2}}{\to }$

(a) PH₃ $\stackrel{\mathrm{w}}{\to }$P +3/2. H₂

This is an example of surface catalysed unimolecular decomposition.

For the above reaction, rate is given as

 Rate = k$\mathrm{\alpha \rho }$/1+$\mathrm{\alpha \rho }$

where, $\mathrm{\rho }$ = partial pressure of absorbing substrate.

At low pressure, $\mathrm{\alpha \rho }$

(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.