Chemical Kinetics, on the exam screen.

The unit of the rate constant of an nth-order reaction is $(\text{mol L}^{-1})^{1-n}\,\text{s}^{-1}$. For n=0 this is $\text{mol L}^{-1}\text{s}^{-1}$ (IV); n=1 gives $\text{s}^{-1}$ (III); n=2 gives $\text{mol}^{-1}\text{L s}^{-1}$ (I); n=3 gives $\text{mol}^{-2}\text{L}^2\text{s}^{-1}$ (II). Hence P-IV, Q-III, R-I, S-II.

A straight-line plot of concentration [R] against time corresponds to a zero-order reaction, for which $[R] = [R]_0 - kt$; the constant (linear) rate of decrease means the rate is independent of concentration and equal to the rate constant k (= -slope). First- and higher-order reactions give curved concentration-time plots, not straight lines.

Comparing with the Arrhenius form $\ln k = \ln A - \frac{E_a}{RT}$, the slope term gives $\frac{E_a}{R}=1.25\times10^4$. Thus $E_a = 1.25\times10^4 \times 1.987 \approx 24837$ cal mol$^{-1}$ $\approx 24.84$ kcal mol$^{-1}$ (1).

The rate of REACTION is defined as $-\frac{1}{2}\frac{d[A]}{dt} = k$ because of the stoichiometric coefficient 2, so A itself disappears at $2k$ per minute. Integrating, $[A] = [A]_0 - 2kt$. Completing 75% means 1.5 M of A has been consumed, so $1.5 = 2(1.0)t$ and $t = 0.75\ \text{min}$. NCERT Class 12, Chemical Kinetics (integrated rate equation for zero order).

In logarithmic form the Arrhenius equation is $\ln k = \ln A - \frac{E_a}{RT}$, a straight line of slope $-E_a/R$ whose intercept on the $\ln k$ axis is $\ln A$. The graph shows that intercept at 6, so $\ln A = 6$. Setting $\ln k = 2$ gives $2 = 6 - \frac{6640}{8.3\,T}$, hence $\frac{6640}{8.3\,T} = 4$ and $T = 200\ \text{K}$. NCERT Class 12, Chemical Kinetics (temperature dependence of rate).

Why the other options are wrong:

- Option (1): at 125 K, $\frac{6640}{8.3 \times 125} = 6.4$, so $\ln k = 6 - 6.4 = -0.4$, not 2
- Option (2): at 150 K, $\frac{6640}{8.3 \times 150} \approx 5.33$, so $\ln k \approx 0.67$, not 2
- Option (4): at 250 K, $\frac{6640}{8.3 \times 250} = 3.2$, so $\ln k = 2.8$, not 2

For a first-order reaction, $t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]} \right)$. Substituting, $t = \frac{2.303}{4.606 \times 10^{-3}} \log \left( \frac{2.0}{0.2} \right) = \frac{2.303}{4.606 \times 10^{-3}} \log 10 = \frac{2.303}{4.606 \times 10^{-3}} \times 1 = 500\ s$. Chapter: Chemical Kinetics.

An increase in the concentration of reactants leads to a higher collision frequency, as more particles are available to collide. This does not affect activation energy, heat of reaction, or threshold energy. Chapter: Chemical Kinetics.