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]
The output (Y) of the given logic gate is similar to the output of an/a :
A | B | Y |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 1 |
1 | 1 | 0 |
List I (Spectral Lines of Hydrogen for transitions from) | List II (Wavelength (nm)) | ||
A. | n2 = 3 to n1 = 2 | I. | 410.2 |
B. | n2 = 4 to n1 = 2 | II. | 434.1 |
C. | n2 = 5 to n1 = 2 | III. | 656.3 |
D. | n2 = 6 to n1 = 2 | IV. | 486.1 |