Question 12 of 37beginner🔧 ApplyNumerical2 marks

In a reaction, 2A→Products2A \rightarrow Products, the concentration of AA decreases from 0.5 mol L−1^{-1} to 0.4 mol L−1^{-1} in 10 minutes. Calculate the rate during this interval?

Correct Answer

The rate of reaction during this interval is 0.005 mol L⁻¹ min⁻¹ (or 5×105 \times 10⁻³ mol L⁻¹ min⁻¹).

Given:

  • Reaction: 2A → Products
  • Initial concentration of A = 0.5 mol L⁻¹
  • Final concentration of A = 0.4 mol L⁻¹
  • Time interval (Δt) = 10 minutes

Step 1: Calculate the change in concentration of A Δ[A] = [A]final - [A]initial = 0.4 - 0.5 = -0.1 mol L⁻¹

Step 2: Calculate the rate of disappearance of A Rate of disappearance of A = -Δ[A]/Δt = -(-0.1)/10 = 0.01 mol L⁻¹ min⁻¹

Step 3: Calculate the rate of reaction For the reaction 2A → Products, the rate is expressed as: Rate = -1/2 × Δ[A]/Δt Rate = -1/2 × (-0.1)/10 = 0.005 mol L⁻¹ min⁻¹

Therefore, the rate during this interval is 0.005 mol L⁻¹ min⁻¹.

Exercise: Intext Questions | Q: 3.2 | (Chapter: 6)
For More Understanding

Explanation

This question tests the student's understanding of expressing the rate of reaction in terms of the rate of disappearance of reactants. For a reaction 2A → Products, the stoichiometric coefficient of A is 2, so the rate must be divided by this coefficient. The rate of reaction is defined as Rate = -1/2 × Δ[A]/Δt, where the negative sign accounts for the decrease in reactant concentration. The context shows similar calculations where rate expressions account for stoichiometric coefficients (e.g., Rate = 1/4 × Δ[NO₂]/Δt for reactions with coefficient 4).

Solution Steps

  1. Step 1: Calculate Δ[A] = 0.4 - 0.5 = -0.1 mol L⁻¹

  2. Step 2: Calculate rate of disappearance of A = -Δ[A]/Δt = -(-0.1)/10 = 0.01 mol L⁻¹ min⁻¹

  3. Step 3: Apply stoichiometric factor: Rate = -1/2 × Δ[A]/Δt = -1/2 × (-0.1)/10 = 0.005 mol L⁻¹ min⁻¹

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