For the reaction, A+B \\toP=-\frac{d[A]}{d t}=-\frac{d[B]}{d t}=k[A][B] and k t={1}/{[A]_{0}-[B]_{0}} \ln {[A][B]_{0}}/{[B][A]_{0}} when, [A]_{0} \\neq[B]_{0} If, [A]_{0}=[B]_{0} then the integrated rate law will be

Correct Answer is : 1[A]=1[A]0+kt or 1[B]=1[B]0+kt

\frac{1}{[B]}={1}/{[A]_{0}}+k t

\frac{1}{[A]}={1}/{[B]_{0}}+k t

\frac{1}{[A]}={1}/{[A]_{0}}+k t or \frac{1}{[B]}={1}/{[B]_{0}}+k t

AMU Engineering 2018 solved paper Solved Paper

Section: Chemistry

Question : 88 of 150

Marks: +1, -0

For the reaction,
A+B→P=−

d[A]

=−

d[B]

=k[A][B] and kt=

[A]0−[B]0

‌ln‌

[A][B]0

[B][A]0

when, [A]0≠[B]0
If, [A]0=[B]0 then the integrated rate law will be

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