For the following elementary reaction: 2Br^● → Br₂. The rate of consumption of the reactant and rate of formation of product is given by which set of expressions?

Question

Quizplus · Accepted Answer

The coefficient of the reactant molecule in the balanced chemical equation is halved compared to that of the product molecule. Therefore, the rate of consumption of the reactant ($\mathrm{d}\left[\mathrm{Br}^{*}\right]/\mathrm{d} t$) should be halved and the rate of formation of the product ($\mathrm{d}\left[\mathrm{Br}_{2}\right]/\mathrm{d} t$) should remain the same. Therefore, the correct set of expressions is $-\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{Br}^{*}\right]}{\mathrm{d} t}=\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{d} t}$.

For the Following Elementary Reaction: 2Br● → Br2 −d[Br∗]dt=d[Br2]dt-\frac{\mathrm{d}\left[\mathrm{Br}^{*}\right]}{\mathrm{d} t}=\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{d} t}−dtd[Br∗]​=dtd[Br2​]​

For the Following Elementary Reaction: 2Br^● → Br₂ $-\frac{\mathrm{d}\left[\mathrm{Br}^{*}\right]}{\mathrm{d} t}=\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{d} t}$