Arrange the following reactions in increasing order of the value of the ratio of rate constants ($k_{310}/k_{300}$)
(A) $A+B → C: E_a = 70\, kJ$
(B) $P+Q→R: E_a = 45\, kJ$
(C) $G+H→I: E_a = 95\, kJ$
(D) $X+Y→Z: E_a = 100\, kJ$
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B), (A), (C), (D)
Core Formula (Arrhenius Ratio):
$\ln \left( \frac{k_{310}}{k_{300}} \right) = \frac{E_a}{R} \left( \frac{1}{300} - \frac{1}{310} \right)$
$\frac{1}{300} - \frac{1}{310} = \frac{10}{93000} = 1.075 \times 10^{-4}$
$R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}$
(A) $E_a = 70 \, \text{kJ} = 70000 \, \text{J}$:
$\ln \left( \frac{k_{310}}{k_{300}} \right) = \frac{70000}{8.314} \times 1.075 \times 10^{-4}$
$= 8421 \times 1.075 \times 10^{-4} = 0.905$
$\frac{k_{310}}{k_{300}} = e^{0.905} \approx 2.47$
(B) $E_a = 45 \text{ kJ} = 45000 \text{ J}$
$\ln\left(\frac{k_{310}}{k_{300}}\right) = \frac{45000}{8.314} \times 1.075 \times 10^{-4}$
$= 5413 \times 1.075 \times 10^{-4} = 0.582$
$\frac{k_{310}}{k_{300}} = e^{0.582} \approx 1.79$
(C) $E_a = 95 \text{ kJ} = 95000 \text{ J}$
$\ln\left(\frac{k_{310}}{k_{300}}\right) = \frac{95000}{8.314} \times 1.075 \times 10^{-4}$
$= 11428 \times 1.075 \times 10^{-4} = 1.229$
$\frac{k_{310}}{k_{300}} = e^{1.229} \approx 3.42$
(D) $E_a = 100 \text{ kJ} = 100000 \text{ J}$
$\ln\left(\frac{k_{310}}{k_{300}}\right) = \frac{100000}{8.314} \times 1.075 \times 10^{-4}$
$= 12027 \times 1.075 \times 10^{-4} = 1.293$
$\frac{k_{310}}{k_{300}} = e^{1.293} \approx 3.64$
Comparison:
|
Reaction |
Ratio $k_{310}/k_{300}$ |
|
B |
1.79 |
|
A |
2.47 |
|
C |
3.42 |
|
D |
3.64 |