For the reaction $2N_2O_5(g) \to 4NO_2(g) + O_2(g)$, if $N_2O_5$ disappears at 0.02 mol dm⁻³ s⁻¹, the **rate of formation of $O_2$** is:
A0.01 mol dm⁻³ s⁻¹
B0.08 mol dm⁻³ s⁻¹
C0.02 mol dm⁻³ s⁻¹
D0.04 mol dm⁻³ s⁻¹
Answer & Solution
Correct answer: A. 0.01 mol dm⁻³ s⁻¹
$-\frac{1}{2}d[N_2O_5]/dt = +d[O_2]/dt$ ⇒ $d[O_2]/dt = (1/2)(0.02) = 0.01$ mol dm⁻³ s⁻¹.
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