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The gas -phase thermal decomposition of one mole of $(CH_3)_3CO_2C(CH_3)_3$ in a constant volume apparatus yields two moles of acetone and one mole of ethane.If the reaction obeys ${1^{st}}$ order then which of the following is true?

$\begin{array}{1 1}(a)\;k_1=\large\frac{1}{t}\normalsize ln\large\frac{P_i}{P_i-P}\\(a)\;k_1=\large\frac{1}{t}\normalsize ln\large\frac{2P_i}{2P_i-P}\\(c)\;k_1=\large\frac{1}{t}\normalsize ln\large\frac{2P_i}{3P_i-P}\\(d)\;k_1=\large\frac{1}{t}\normalsize ln\large\frac{P_i}{3P_i-P}\end{array}$

1 Answer

$(CH_3)_3CO_2C(CH_3)_3\rightarrow 2(CH_3)_2C=O+C_2H_6$
Let initial pressure of compound be $P_i$.If $x$ is decrease in pressure after time t then pressure after time t is $(p_i-x)$ the acetone have 2x and ethane has $x.$
So $P=(P_i-x)+2x+x=P_i+2x$
Pressure of reactant =$P_i-x=P_i-\large\frac{P-P_i}{2}$
$\Rightarrow \large\frac{3P_i-P}{2}$
Hence $k_1=\large\frac{1}{t}$$ln\large\frac{a}{a-x}=\frac{1}{t}$$ln\large\frac{P_i}{(3P_i-P)/2}$
$\Rightarrow \large\frac{1}{t}$$ln\large\frac{2P_i}{(3P_i-P)}$
Hence (c) is the correct answer.
answered Dec 19, 2013 by sreemathi.v

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