For the reaction, 2A → B, rates= k[A]2. If the concentration of reactant is doubled, then the:
(a) | rate of reaction will be doubled. |
(b) | rate constant will remain unchanged, however rate of reaction is directly proportional to the rate constant. |
(c) | rate constant will change since the rate of reaction and rate constant are directly proportional to each other. |
(d) | rate of reaction will increase by four times. |
Identify the set of correct statements & choose the correct answer from the options given below:
1. | (a) and (c) only | 2. | (a) and (b) only |
3. | (b) and (d) only | 4. | (c) and (d) only |
For a general reaction A → B, the plot of concentration of A vs time is given below:
The unit of the rate constant would be:
1. | \(\mathrm{mol}^{-1} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \) | 2. | \(s^{-1} \) |
3. | \(\mathrm{mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \) | 4. | None of the above |
For a general reaction A → B, the plot of concentration of A vs time is given below:
The order of the reaction would be-
1. Zero
2. First
3. Half
4. Second
The correct statement about X in the below mentioned graph:
1. | X represents activation energy without catalyst. |
2. | X represents activation energy with catalyst. |
3. | X represents the enthalpy of the reaction without a catalyst. |
4. | X represents the enthalpy of the reaction with a catalyst. |
The graph between lnK and 1/T is given below:
The value of activation energy would be:
1. | \(207.8\ \mathrm{KJ} / \mathrm{mol} \) | 2. | \(- 207.8\ \mathrm{KJ} / \mathrm{mol} \) |
3. | \(210.8\ \mathrm{KJ} / \mathrm{mol} \) | 4. | \(-210.8\ \mathrm{KJ} / \mathrm{mol} \) |
Consider the following graph:
The point that represents the most probable kinetic energy is -
1. C
2. A
3. B
4. None of the above
The nature of the reaction represented in the following graph is:
1. | Endothermic reaction |
2. | Exothermic reaction |
3. | Both endothermic and exothermic reactions are represented by the same graph. |
4. | None of the above |
Consider the following graph:
The instantaneous rate of reaction at t = 600 sec will be:
Based on the graph below, the average rate of reaction will be:
1. \(\frac{[R_{2}]-[R_{1}]}{t_{2}-t_{1}}\)
2. \(-(\frac{[R_{2}]-[R_{1}]}{t_{2}-t_{1}})\)
3. \(\frac{[R_{2}]}{t_{2}}\)
4. \(-(\frac{[R_{1}]-[R_{2}]}{t_{2}-t_{1}})\)