For the reaction 3,{A_{,(g),}},xrightarr | vedclass

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Question

$K=10^{-14}$  $L/mol.\min$

$-\frac{1}{3} \frac{\mathrm{d}(\mathrm{A})}{\mathrm{dt}}=\mathrm{K}(\mathrm{A})^{2}$

$-\frac{\mathrm{d}(\mathrm{

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$K=10^{-14}$  $L/mol.\min$

$-\frac{1}{3} \frac{\mathrm{d}(\mathrm{A})}{\mathrm{dt}}=\mathrm{K}(\mathrm{A})^{2}$

$-\frac{\mathrm{d}(\mathrm{A})}{\mathrm{dt}}=3 	imes 10^{-14} 	imes(0.5)^{2}$

$ = 7.5 	imes {10^{ - 15}}\,\,{	ext{M}}/\min $

$=\frac{7.5 	imes 10^{-15}}{60}=1.25 	imes 10^{-16}\, \mathrm{M} / \mathrm{sec}$

For the reaction $3\,{A_{\,(g)\,}}\,\xrightarrow{K}\,{B_{(g)}}\, + \,\,{C_{(g)\,,}}K$ is ${10^{ - 14}}\,L/mol.\min .$ if $(A) = 0.5\,M$ then the value of $ - \frac{{d(A)}}{{dt}}$ (in $M / sec$ ) is.

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