For the reaction $3I^-(aq) + S_2O_8^{2-}(aq) \to I_3^-(aq) + 2SO_4^{2-}(aq)$, if the rate of formation of $SO_4^{2-}$ is 0.022 mol dm⁻³ s⁻¹, the rate of consumption of $I^-$ is:
A0.011 mol dm⁻³ s⁻¹
B0.033 mol dm⁻³ s⁻¹
C0.022 mol dm⁻³ s⁻¹
D0.066 mol dm⁻³ s⁻¹
Answer & Solution
Correct answer: B. 0.033 mol dm⁻³ s⁻¹
Overall rate: $-\frac{1}{3}d[I^-]/dt = +\frac{1}{2}d[SO_4^{2-}]/dt$. So $d[I^-]/dt = -(3/2)(0.022) = -0.033$ mol dm⁻³ s⁻¹, i.e. $I^-$ is consumed at 0.033 mol dm⁻³ s⁻¹.
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