A-Level Chemistry May/June 2025 Q2(a)(ii): In experiment 1, the initial concentrations of NO and Cl2 are equal. The initial rate o…

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  From the rate equation, Rate = $k$[NO]$^2$[Cl$_2$]. The value of the rate constant, $k$, is 26.4 dm$^6$mol$^{-2}$s$^{-1}$.

  Given that Rate = $2.57 \times 10^{-6}$ mol dm$^{-3}$s$^{-1}$ and [NO] = [Cl$_2$]. Let [NO] = $x$, therefore [Cl$_2$] = $x$.

  Substitute values into the rate equation: $2.57 \times 10^{-6} = 26.4 \times (x)^2 \times (x)$ $2.57 \times 10^{-6} = 26.4 \times x^3$

  Rearrange to solve for $x^3$: $x^3 = \frac{2.57 \times 10^{-6}}{26.4}$ $x^3 = 9.7348... \times 10^{-8}$

  Take the cube root to find $x$: $x = \sqrt[3]{9.7348... \times 10^{-8}}$ $x = 4.600... \times 10^{-3}$

  Initial concentration of NO = $4.60 \times 10^{-3}$ mol dm$^{-3}$ (to 3 s.f.)

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  How the marks are awarded

  • M1 — Correctly substituting the rate and concentrations into the rate equation, recognising that [NO]$^2$[Cl$_2$] becomes [concentration]$^3$, to set up the expression: $2.57 \times 10^{-6} = 26.4 \times [\text{concentration}]^3$.
  • M2 — Correctly calculating the final concentration to 3 significant figures, giving the answer $4.60 \times 10^{-3}$.

  Common mistakes

  • Forgetting to combine the concentration terms, leading to an algebraic error (e.g., using [conc]$^2$ instead of [conc]$^3$).
  • Making a calculation error when rearranging, such as forgetting to take the cube root of the result, or taking the square root by mistake.
  • Incorrectly rounding the final answer or not giving it to the required 3 significant figures.
  • Using the stoichiometry of the overall equation (2NO) to incorrectly assume [NO] = 2[Cl$_2$], despite the question stating they are equal.

  Examiner tip: When calculating concentration from a rate equation, be methodical in your algebraic rearrangement and ensure you apply the correct root (e.g., cube root for overall order 3) as the final step.

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