Question
Chemistry Question on Chemical Kinetics
A chemical reaction proceeds into the following steps Step I,\hspace15mm 2A \rightleftharpoons X fast Step II,\hspace10mm X + B \rightleftharpoons Y slow Step III,\hspace10mm Y + B \rightleftharpoons Product fast The law for the overall reaction is
A
rate=k[A]2
B
rate=k[B]2
C
rate=k[A][B]
D
rate=k[A]2[B]
Answer
rate=k[A]2[B]
Explanation
Solution
dtdx=k′[X][B]
From step I \hspace10mm K_{eq} = \frac{[X]}{[A]^2}
\hspace25mm [X] = K_{eq} [A]^2
\hspace25mm \frac{dx}{dt} = k' . K_{eq} [A]^2[B]
\hspace25mm = k [A]^2[B]