For the same reaction ($2N_2O_5 \to 4NO_2 + O_2$, $d[N_2O_5]/dt = -0.02$), the **rate of formation of $NO_2$** is:
A0.08 mol dm⁻³ s⁻¹
B0.02 mol dm⁻³ s⁻¹
C0.04 mol dm⁻³ s⁻¹
D0.01 mol dm⁻³ s⁻¹
Answer & Solution
Correct answer: C. 0.04 mol dm⁻³ s⁻¹
$-\frac{1}{2}d[N_2O_5]/dt = \frac{1}{4}d[NO_2]/dt$ ⇒ $d[NO_2]/dt = 2\,|d[N_2O_5]/dt| = 2(0.02) = 0.04$ mol dm⁻³ s⁻¹.
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