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Q.

The reaction $v_{1} A + v_{2} B ⟶$ products is first-order with respect to A and zero-order with respect to B. If the reaction is started with ${[A]}_{0}$ and ${[B]}_{0}$ , the integrated rate expression of this reaction would be

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a

$\ln \frac{[A]_{0}}{[A]_{0} - v_{1} x} = v_{1} k_{1} t$

b

$\ln \frac{[A]_{0}}{[A]_{0} - v_{1} x} = v_{1} k_{1} t$

where x is the extent of reaction divided by constant volume.

c

$\ln \frac{[A]_{0}}{[A]_{0} - v_{1} x} = k_{1} t$

d

$\ln \frac{[A]_{0}}{[A]_{0} - x} = k_{1} t$

answer is C.

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Detailed Solution

We have

$- \frac{1}{v_{1}} \frac{d [A]}{d t} = k [A] i.e. \int_{[A]_{]}}^{[A]} \frac{d [A]}{[A]} = - v_{1} k \int_{0}^{l} d t$

$\ln \frac{[A]}{[A]_{0}} = - v_{1} k t or \ln \frac{[A]_{0}}{[A]_{0} - v_{1} x} = v_{1} k t$

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