MHT CET · Chemistry · Chemical Kinetics
For a reaction \(\mathrm{A}+\mathrm{B} \rightarrow\) product, if \([\mathrm{A}]\) is doubled keeping [B] constant, the rate of reaction doubles. Calculate the order of reaction with respect to \(A\).
- A \(0\)
- B \(1 / 2\)
- C \(1\)
- D \(2\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \mathrm{A}+\mathrm{B} \longrightarrow \text { Product } \\ & \text { Rate }^{\mathrm{k}[\mathrm{A}]^x[\mathrm{~B}]^{\mathrm{y}}} \\ & \text { Rate }_1=\mathrm{k}[2 \mathrm{~A}]^x[\mathrm{~B}]^{\mathrm{y}} \\ \therefore \quad & \frac{\text { Rate }_1}{\text { Rate }}=\frac{\mathrm{k} 2^x[\mathrm{~A}]^x[\mathrm{~B}]^y}{\mathrm{k}[\mathrm{A}]^x[\mathrm{~B}]^{\mathrm{y}}}\end{array}\)
\(\begin{array}{ll}
\therefore & \frac{2 \times \text { Rate }}{\text { Rate }}=2^x \\
\therefore & 2^x=2 \\
\therefore & x=1
\end{array}\)
\(\therefore \quad\) Order of reaction with respect to \(A=1\)
\(\begin{array}{ll}
\therefore & \frac{2 \times \text { Rate }}{\text { Rate }}=2^x \\
\therefore & 2^x=2 \\
\therefore & x=1
\end{array}\)
\(\therefore \quad\) Order of reaction with respect to \(A=1\)
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