Question

For a first order reaction $A\left(g\right)\to 2B\left(g\right)+C\left(g\right)$ at constant volume and 300 K, the total pressure at the beginning (t=0) and time t are $P_o$ and $P_t$ respectively. Initially, only $A$ is present with concentration $[A{]}_o$ and $t_{1/2}$ is the time required for the partial pressure of $A$ to reach a third of its initial value. The correct options are:
(Assume that all these gases behave as ideal gaes)

• A.
• B.
• C.
• D.

Answer 1

Answer: (A), (D)

Since it is a first order reaction wrt A, we know that Rate constant will be independent of initial concentration. Hence, the plot will be a straight line parallel to the X-axis.
$A\left(g\right)\to 2B\left(g\right)+C\left(g\right)$
t=0 : $P_{o}00$
t=t : $P_o-P2PP$
$P_t=P_o-P+2P+P=P_o-2P$
$⟹P=\frac{P_t-P_o}{2}$
$⟹P_o-P=\frac{3P_o-P_t}{2}$
Substituting in the integrated rate law, we get:
$kt=ln\left(\frac{P_{A_o}}{P_{A_t}}\right)=ln\frac{2P_o}{3P_o-P_t}$
$kt=ln2P_o-ln(3P_o-P_t)$
Rearranging, we get: $ln(3P_o-P_t)=-kt+ln2P_o$
This is in the standard slope format $y=mx+c$
Hence, we can see how this also corresponds to the graph.