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The folllowing data were obtained during the first order decomposition of $2 A(g) \rightarrow B(g)+C(s)$ at a constant volume and at a particular temperature
The rate constant in $\mathrm{min}^{-1}$ is
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The rate constant in $\mathrm{min}^{-1}$ is
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Verified Answer
The correct answer is:
$0.0693$
$$
\begin{array}{lc}
2 A(g) \longrightarrow B(g)+C(s) \\
2-2 x & x
\end{array}
$$
At the end of reaction, only 1 mole of gas is present whose pressure is 200 Pascal.
$\therefore$ At the beginning of the reaction 2 moles of gas should have a pressure of 400 Pascal.
After time $10 \mathrm{~min}$
No. of moles present,
$$
2-2 x+x=2-x
$$
The pressure of 2 moles $=400$
$\therefore \quad 400-x=300$
$\therefore \quad x=100$
$\therefore$ Pressure due to $2-2 x$ moles of $A$
$$
=400-200=200
$$
$\begin{aligned} \therefore \quad k &=\frac{2.303}{t} \log \left(\frac{a}{a-x}\right)=\frac{2.303}{10} \log \left(\frac{400}{200}\right) \\ &=\frac{2.303}{10} \log 2=\frac{0.693}{10} \\ &=0.0693 \mathrm{~min}^{-1} \end{aligned}$
\begin{array}{lc}
2 A(g) \longrightarrow B(g)+C(s) \\
2-2 x & x
\end{array}
$$
At the end of reaction, only 1 mole of gas is present whose pressure is 200 Pascal.
$\therefore$ At the beginning of the reaction 2 moles of gas should have a pressure of 400 Pascal.
After time $10 \mathrm{~min}$
No. of moles present,
$$
2-2 x+x=2-x
$$
The pressure of 2 moles $=400$
$\therefore \quad 400-x=300$
$\therefore \quad x=100$
$\therefore$ Pressure due to $2-2 x$ moles of $A$
$$
=400-200=200
$$
$\begin{aligned} \therefore \quad k &=\frac{2.303}{t} \log \left(\frac{a}{a-x}\right)=\frac{2.303}{10} \log \left(\frac{400}{200}\right) \\ &=\frac{2.303}{10} \log 2=\frac{0.693}{10} \\ &=0.0693 \mathrm{~min}^{-1} \end{aligned}$
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