The decomposition of sucrose follows the first-order rate law. For this decomposition, t1/2 is 3.00 hours. The fraction of a sample of sucrose that remains after 8 hours would be:
1. | 0.13 | 2. | 0.42 |
3. | 0.16 | 4. | 0.25 |
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
For a reaction A → Product, with k = 2.0 × 10–2 s–1, if the initial concentration of A is 1.0 mol L–1, the concentration of A after 100 seconds would be :
1. | 0.23 mol L–1 | 2. | 0.18 mol L–1 |
3. | 0.11 mol L–1 | 4. | 0.13 mol L–1 |
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.
A first-order reaction takes 40 min for 30 % decomposition. The half life of the reaction will be:
1. | 88.8 min | 2. | 94.3 min |
3. | 67.2 min | 4. | 77.7 min |
The role of a catalyst is to change:
1. | Gibbs energy of the reaction |
2. | Enthalpy of reaction |
3. | The activation energy of the reaction |
4. | Equilibrium constant |
In the presence of a catalyst, the heat evolved or absorbed during the reaction:
1. | Increases. | 2. | Decreases. |
3. | Remains unchanged. | 4. | May increase or decrease. |
1. | Determining the rate constant at standard temperature. |
2. | Determining the rate constant at two temperatures. |
3. | Determining probability of collision. |
4. | Using the catalyst. |
The correct statement based on the graph below is:
1. | The activation energy of the forward reaction is E1 + E2 and the product is less stable than reactant. |
2. | The activation energy of the forward reaction is E1 + E2 and the product is more stable than the reactant. |
3. | The activation energy of both forward and backward reaction is E1 + E2 and reactant is more stable than the product. |
4. | The activation energy of the backward reaction is E1 and the product is more stable than reactant. |