A reactant A covers a catalyst surface completely, so…

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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?

ChemistryAdvancedHigh School (11–12)Chemical Kinetics › Integrated Rate Equations and Half-Life JEE Advanced Olympiad

A reactant A covers a catalyst surface completely, so it turns into B at a steady rate. This step is zero order, with $k_0 = 2.0 \times 10^{-4}\ \text{mol L}^{-1}\ \text{s}^{-1}$ , 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 $k_1 = 4.0 \times 10^{-3}\ \text{s}^{-1}$ . The solution is well stirred and its volume stays fixed. At the start there is no B, and a chemist follows $[\text{B}]$ with a spectrometer. So $\dfrac{d[\text{B}]}{dt} = k_0 - k_1[\text{B}]$ . Which statement about $[\text{B}]$ is correct?

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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?Match each observation in List-I with the overall order of reaction in List-II that it shows. List-I (P) For a reaction of one reactant, the half-life falls from 40.0 min to 20.0 min when the starting concentration is doubled. (Q) The rate constant has the unit L^(1/2) mol^(-1/2) s⁻¹. (R) For rate = k[A]ˣ[B]ʸ, doubling [A] alone makes the reaction 4 times faster, and doubling both [A] and [B] makes it 8 times faster. (S) For a reaction of one reactant, 75 per cent is used up in exactly twice the time taken for 50 per cent, whatever the starting concentration. List-II (1) 0 (2) 1 (3) 3/2 (4) 2 (5) 3 →

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