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# For the reaction $A\rightarrow B$ ,the rate law expression is $-\large\frac{d[A]}{dt}$$=k[A]^{3/2}. If the initial concentration of A is [A]_0, which of the following is true? \begin{array}{1 1}(a)\;k=\large\frac{2}{t}\normalsize(A_o^{-1/2}-A^{-1/2})\\(b)\;\;K=\large\frac{2}{t}\normalsize(A_o^{1/2}-A^{1/2})\\(c)\;t_{1/2}=\large\frac{k}{2[A_o]^{1/2}}\\(d)\;t_{1/2}=\large\frac{ln\;2}{k}\end{array} Can you answer this question? ## 1 Answer 0 votes Answer: t_{1/2}=\large\frac{2}{k}\bigg(\frac{2^{1/2}-1}{[A_o]^{1/2}}\bigg) By differential rate law, \large\frac{d[A]}{dt}=$$-k[A]^{3/2}$
Integrating, $\large \int_{[A_0]}^{[A_t]}$$\large\frac{d[A]}{[A]^3/2}$$=-\large\int_0^t $$kdt \Rightarrow 2([A_o]^{-1/2}-[A_t]^{1/2})=kt k=\large\frac{2}{t}$$((A_o)^{-1/2}-(A_t)^{-1/2})$
Simplifying for $t = t_{1/2}=\large\frac{2}{k}\bigg(\frac{2^{1/2}-1}{[A_o]^{1/2}}\bigg)$
edited Jul 26, 2014