The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation.
Given R = 8.314 J K-1 mol-1, log 4 = 0.6021
Solution:
We can use the Arrhenius equation to solve this problem:
$$ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) $$
Where:
Given:
Substituting the values into the Arrhenius equation:
$$ \ln(4) = \frac{E_a}{8.314}\left(\frac{1}{300.15} - \frac{1}{330.15}\right) $$
We know that ln(4) = 2.303 * log(4) = 2.303 * 0.6021 = 1.3862963
$$ 1.3862963 = \frac{E_a}{8.314}\left(\frac{330.15 - 300.15}{300.15 \times 330.15}\right) $$
$$ 1.3862963 = \frac{E_a}{8.314}\left(\frac{30}{99099.225}\right) $$
$$ 1.3862963 = \frac{E_a}{8.314}(3.02727 \times 10^{-4}) $$
$$ E_a = \frac{1.3862963 \times 8.314}{3.02727 \times 10^{-4}} $$
$$ E_a = \frac{11.5257}{3.02727 \times 10^{-4}} $$
$$ E_a = 38073.47 \text{ J/mol} $$
$$ E_a = 38.07347 \text{ kJ/mol} $$
Therefore, the activation energy is approximately 38.04 kJ/mol.
Option 1: 38.04 KJ/mol
For a reaction, \[ {N}_2{O}_5(g) \rightarrow 2{NO}_2(g) + \frac{1}{2} {O}_2(g) \] in a constant volume container, no products were present initially. The final pressure of the system when 50% of the reaction gets completed is:
In Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl. The percentage composition of chlorine in the compound is ___________%. [Given: Molar mass in g mol\(^{-1}\) of Ag = 108, Cl = 35.5]
List I | List II | ||
A | Down’s syndrome | I | 11th chormosome |
B | α-Thalassemia | II | ‘X’ chromosome |
C | β-Thalassemia | III | 21st chromosome |
D | Klinefelter’s syndrome | IV | 16th chromosome |
The velocity (v) - time (t) plot of the motion of a body is shown below :
The acceleration (a) - time(t) graph that best suits this motion is :