The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5 s–1 at 546 K. If the energy of activation is 179.9 kJ/mol,
the value of the pre-exponential factor will be:
1.
2.
3.
4.
The decomposition of hydrocarbons follows the equation: k = (4.5 × 1011s–1) e–28000K/T
The activation energy (Ea) for the reaction would be:
1. 232.79 kJ mol–1
2. 245.86 kJ mol–1
3. 126.12 kJ mol–1
4. 242.51 kJ mol–1
The rate constant for the first-order decomposition of H2O2 is given by the equation:
\(log \ k \ = \ 14.34 \ - \ 1.25 \ \times \ 10^{4}\frac{K}{T}\).
The value of Ea for the reaction would be:
1. 249.34 kJ mol–1
2. 242.64 J mol–1
3. –275.68 kJ mol–1
4. 239.34 kJ mol–1
The decomposition of A into the product has a value of k as 4.5 × 103 s–1 at 10°C and energy of activation of 60 kJ mol–1. The temperature at which the rate constant becomes 1.5 × 104 s–1 would be -
1. \(24 \text { K} \)
2. \(24 ^\circ \text { C} \)
3. \(31 ^\circ \text { C} \)
4. \(38 ^\circ \text { C} \)
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. The energy of activation of the reaction would be -
1. 65.93 kJ mol–1
2. 52.85 kJ mol–1
3. 55.46 kJ mol–1
4. 60.93 kJ mol–1
The correct statement about the rate constant of a reaction is:
1. | Rate constant is nearly doubled with a rise in temperature by 10 °C |
2. | Rate constant becomes half with a rise in temperature by 10 °C |
3. | Rate constant remains unchanged with a rise in temperature by 10 °C |
4. | None of the above |
A first-order reaction's 10 percent completion time at 298 K is the same as its 25 percent completion time at 308 K. The value of will be:
1.
2.
3.
4.