MHT CET · Chemistry · Chemical Kinetics
For the reaction : \(\mathrm{H}_{2}+\mathrm{I}_{2} \longrightarrow 2 \mathrm{HI}\), the
differential rate law is
- A \(-\frac{d\left[\mathrm{H}_{2}\right]}{d t}=-\frac{d\left[\mathrm{I}_{2}\right]}{d t}=2 \frac{d[\mathrm{HI}]}{d t}\)
- B \(-\frac{d\left[\mathrm{H}_{2}\right]}{d t}=-2 \frac{d\left[\mathrm{I}_{2}\right]}{d t}=\frac{d[\mathrm{HI}]}{d t}\)
- C \(-\frac{d\left[\mathrm{H}_{2}\right]}{d t}=-\frac{d\left[\mathrm{I}_{2}\right]}{d t}=\frac{d[\mathrm{HI}]}{d t}\)
- D \(-\frac{d\left[\mathrm{H}_{2}\right]}{d t}=-\frac{d\left[\mathrm{I}_{2}\right]}{d t}=\frac{d[\mathrm{HI}]}{d t}\)
Answer & Solution
Correct Answer
(B) \(-\frac{d\left[\mathrm{H}_{2}\right]}{d t}=-2 \frac{d\left[\mathrm{I}_{2}\right]}{d t}=\frac{d[\mathrm{HI}]}{d t}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{H}_{2}+\mathrm{I}_{2} \longrightarrow 2 \mathrm{HI}\)
Rate of reaction \(=\frac{-d\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}=\frac{-d\left[\mathrm{I}_{2}\right]}{\mathrm{dt}}=\frac{1}{2} \frac{d[\mathrm{HI}]}{\mathrm{d} t}\)
or \(\quad \frac{-2 d\left[\mathrm{H}_{2}\right]}{d t}=\frac{-2 d\left[\mathrm{I}_{2}\right]}{d t}=\frac{d[\mathrm{HI}]}{d t}\)
Rate of reaction \(=\frac{-d\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}=\frac{-d\left[\mathrm{I}_{2}\right]}{\mathrm{dt}}=\frac{1}{2} \frac{d[\mathrm{HI}]}{\mathrm{d} t}\)
or \(\quad \frac{-2 d\left[\mathrm{H}_{2}\right]}{d t}=\frac{-2 d\left[\mathrm{I}_{2}\right]}{d t}=\frac{d[\mathrm{HI}]}{d t}\)
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