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For a reaction $$2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$$. The rate of consumption of $$\text{SO}_2$$ is $$4 \times 10^{-2}$$ moles/lit/sec. What is the rate of consumption of $$\text{O}_2$$ (in moles/lit/sec) is
Correct Answer: 0.02
By the stoichiometry, $$\text{O}_2$$ is consumed at half the rate of $$\text{SO}_2$$, since $$-\frac{1}{2}\frac{d[\text{SO}_2]}{dt} = -\frac{d[\text{O}_2]}{dt}$$. So the rate of consumption of $$\text{O}_2$$ is $$\frac{4 \times 10^{-2}}{2} = 2 \times 10^{-2} = 0.02$$ moles/lit/sec.
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