Q. No.The rate Constant of reaction A B is 0.6 × 10–3 mole per second. If the Concentration of A is 5, then the concentration of B after 20 min is:
1. 1.08M
2. 3.60M
3. 0.36M
4. 0.72M
2. ln vs T
3. ln K vs
4. ln
When the initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction will be:
1. 0
2. 1
3. 1.5
4. 2
(R = 8.314 J mol–1 K–1)
2. 34.7 kJ mol–1
3. 15.1 kJ mol–1
4. 342 kJ mol–1
A reaction having equal energies of activation for forward and reverse reaction has:
1. ΔG = 0
2. ΔH = 0
3. ΔH = ΔG = ΔS = 0
4. ΔS = 0
| 1. | 12 | 2. | 16 |
| 3. | 32 | 4. | 10 |
| 1. | 1 | 2. | 2 |
| 3. | 3 | 4. | 0 |
In a reaction, A + B → Product, the rate is doubled when the concentration of B is doubled, and the rate increases by a factor of 8, when the concentrations of both the reactants (A and B) are doubled. The rate law for the reaction can be written as:
1. Rate = k[A][B]2
2. Rate = k[A]2[B]2
3. Rate = k[A][B]
4. Rate = k[A]2[B]
In a zero-order reaction for every 10 °C rise of temperature, the rate is doubled.
If the temperature is increased from 10 °C to 100 °C, the rate of the reaction will become:
1. 256 times
2. 512 times
3. 64 times
4. 128 times
Activation energy and rate constant (k1 and k2) of a chemical reaction at two different temperatures (T1 and T2) are related by:
| 1. | \(\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\) |
| 2. | \(\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}+\frac{1}{T_1}\right)\) |
| 3. | \(\ln \frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\) |
| 4. | \(\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\) |
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