A compound in the gas phase can break down by two first…

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A compound in the gas phase can break down by two first order paths, X and Y, that give different products. For path X, A = 1.0 × 10¹³ s⁻¹ and Ea = 120 kJ mol⁻¹. For path Y, A = 1.0 × 10¹⁰ s⁻¹ and Ea = 90 kJ mol⁻¹. Assume that these values do not change with temperature. A chemist wants to know when the two paths give equal amounts of product, and which path takes over as the gas is heated further. Choosing the temperature is a simple way to steer a reaction towards the product that is wanted. Take R = 8.314 J K⁻¹ mol⁻¹. Which statement is correct?

ChemistryAdvancedHigh School (11–12)Chemical Kinetics › Temperature, the Arrhenius Equation and Collision Theory JEE Advanced Olympiad

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A compound in the gas phase can break down by two first order paths, X and Y, that give different products. For path X, $A = 1.0 \times 10^{13}\ \text{s}^{-1}$ and $E_a = 120\ \text{kJ mol}^{-1}$ . For path Y, $A = 1.0 \times 10^{10}\ \text{s}^{-1}$ and $E_a = 90\ \text{kJ mol}^{-1}$ . Assume that these values do not change with temperature. A chemist wants to know when the two paths give equal amounts of product, and which path takes over as the gas is heated further. Choosing the temperature is a simple way to steer a reaction towards the product that is wanted. Take $R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}$ . Which statement is correct?

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← The figure shows the energy profile of a reaction with and without a catalyst. Without the catalyst the activation energy is 75.0 kJ mol⁻¹. With it, the activation energy is 50.0 kJ mol⁻¹. The enthalpy change of the reaction is -30.0 kJ mol⁻¹. Assume that the catalyst leaves the pre-exponential factor unchanged, and that Ea and A do not depend on temperature. The catalysed reaction runs at 300 K. To what temperature T must the uncatalysed reaction be heated to go just as fast, from the same concentrations? And what is the activation energy of the reverse reaction on the catalysed path? Which choice is correct?A reactant A covers a catalyst surface completely, so it turns into B at a steady rate. This step is zero order, with k₀ = 2.0 × 10⁻⁴ mol L⁻¹ s⁻¹, and A is in such excess that the rate stays the same for the whole run. B itself breaks down in the solution by a first order reaction with k₁ = 4.0 × 10⁻³ s⁻¹. The solution is well stirred and its volume stays fixed. At the start there is no B, and a chemist follows [B] with a spectrometer. So (d[B])/dt = k₀ - k₁[B]. Which statement about [B] is correct? →

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