For a first order reaction Ag - 2Bg -Cg at constant volume and 300 K- the total pressure at the beginning t- 0 and at time t are P0 and Pt- respectively- Initially- only A is present with concentration -A0- and t1-3 is the time required for the partial pressure of A to reach 1-3rd of its initial vlaue- The correct options is are Assume that all these gases behave as ideal gases

At t =0 A(g)→ amp; nbsp;2B(g) + amp; nbsp;C(g) p_0 amp; 0 amp; 0 (p_0 p) amp; 2p amp; p p_t = p_0 p + 2p + p = p_0 + 2p p =(p_t ...

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Standard XII

Chemistry

First Order Reaction

For a first o...

Question

For a first order reaction $A (g) \to 2 B (g) + C (g)$ at constant volume and 300 K, the total pressure at the beginning (t= 0) and at time t are $P_{0}$ and $P_{t}$ , respectively. Initially, only A is present with concentration $[A_{0}]$ , and $t_{\frac{1}{3}}$ is the time required for the partial pressure of A to reach ${\frac{1}{3}}^{r d}$ of its initial vlaue. The correct option(s) is (are) (Assume that all these gases behave as ideal gases)

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Solution

The correct option is D
at $t = 0 \begin{matrix} A (g) \to & 2 B (g) + & C (g) p_{0} & 0 & 0 (p_{0} - p) & 2 p & p \end{matrix}$
$p_{t} = p_{0} - p + 2 p + p = p_{0} + 2 p$
$p = \frac{(p_{t} - p_{0})}{2}$
$t = \frac{1}{K} l n \frac{2 p_{0}}{3 p_{0} - p_{t}}$
$k t = l n 2 p_{0} - l n (3 p_{0} - p_{t})$
$l n (3 p_{0} - p_{t}) = l n 2 p_{0} - k t$
This is analogous to $y = m x + c$
So, slope (m) = -k i.e. negative slope.
$c = l n (2 p_{0})$

Again,
$k t_{\frac{1}{3}} = l n \frac{p_{0}}{\frac{p_{0}}{3}}$
$t_{\frac{1}{3}} = \frac{l n 3}{k}$

Here rate constant doest not depend on concentration of A, since we are considering in gas phase. So in the plot of rate constant vs $[A]_{o}$ we will get a parallel line of X-axis.

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