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| 3. | 2 | 4. | 8 |
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| 1. | Increase the forward reaction rate. |
| 2. | Shift the equilibrium to favor the products. |
| 3. | Alter the reaction pathway. |
| 4. | Increase the speed at which equilibrium will be achieved. |
The Arrhenius equations for two reactions are:
A → B: k₁ = 10⁸ exp(−600/RT)
X → Y: k₂ = 10¹⁰ exp(−1800/RT)
The temperature (in Kelvin) at which k₁ = k₂ is:
(Given R = 2 cal mol⁻¹ K⁻¹)
| 1. | \(1200 \mathrm{~K}\) | 2. | \(1200 \times 4.606 \mathrm{~K}\) |
| 3. | \(\dfrac{1200}{4.606} \mathrm{~K}\) | 4. | \(\dfrac{600}{4.606} \mathrm{~K}\) |
The forward rate constant for an elementary reversible gaseous reaction is given as
\(\mathrm{C}_2 \mathrm{H}_6 \rightleftharpoons 2 \mathrm{CH}_3 \text { is } 1.57 \times 10^{-3} \mathrm{~s}^{-1} \text { at } 100 \mathrm{~K}\)
What is the rate constant for the backward reaction at this temperature if \(10^{-4}\) moles of \(\mathrm{CH}_3\) and \(10\) moles of \(\mathrm{C}_2 \mathrm{H}_6\) are present in a \(10\) litre vessel at equilibrium?
1. \(1.57 \times 10^9 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\)
2. \(1.57 \times 10^{10} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\)
3. \(1.57 \times 10^{11} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\)
4. \(1.57 \times 10^7 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\)
| (i) | A catalyst lowers the activation energy of a reaction. |
| (ii) | A catalyst allows the same rate of reaction to be achieved at a lower temperature. |
| (iii) | A catalyst mixes with the reactants and increases the overall concentration of reactants in the rate equation. |
| 1. | A second-order reaction is always a multistep reaction. |
| 2. | A zero-order reaction is a multistep reaction. |
| 3. | A first-order reaction is always a single-step reaction. |
| 4. | A zero-order reaction is a single-step reaction. |