Question

For the first order homogeneous gaseous reaction $A\to 2B+C$, the initial pressure was $P_i$ while total pressure at time 't' was $P_t$. Write expression for the rate constant k in terms of $P_i,P_t$ & $t$.

• A. $k=\frac{2.303}{t}log\left(\frac{2P_i}{3P_i-P_t}\right)$
• B. $k=\frac{2.303}{t}log\left(\frac{2P_i}{2P_t-P_i}\right)$
• C. $k=\frac{2.303}{t}log\left(\frac{P_i}{P_i-P_t}\right)$
• D. None of these

Answer 1

The correct option is A $k=\frac{2.303}{t}log\left(\frac{2P_i}{3P_i-P_t}\right)$

The first-order reaction is $A\to 2B+C$

	A	B	C
At time 0:	PiPi	0	0
At time t:	Pi−xPi−x	2x2x	xx

Hence, the expression for the rate constant will be:

$K=\frac{2.303}{t}log\left(\frac{P_i}{P_i-x}\right)$

But $P_i-x+2x+x=P_t$

Hence, $x=\frac{1}{2}(P_t-P_i)$ or $P_i-x=\frac{1}{2}(3P_i-P_t)$

Thus the expression for the rate constant becomes

$K=\frac{2.303}{t}log\left(\frac{2P_i}{(3P_i-P_t)}\right)$

Hence, the correct option is $\text{A}$