Question

The reaction $A\left(s\right)\to 2B\left(g\right)+C\left(g\right)$ is first order. The pressure after $20$ min. and after a very long time are $150$ mm $Hg$ and $225$ mm $Hg$. The value of rate constant and pressure after $40$ min. are?

• A. $0.05$ $In$ $1.5$ $mi{n}^{-1}$, $200$ mm
• B. $0.5$ $In2$ $mi{n}^{-1}$, $300mm$
• C. $0.05$ $In3$ $mi{n}^{-1}$, $300mm$
• D. $0.05In3$ $mi{n}^{-1}$, $200mm$

Answer 1

The correct option is C $0.05$ $In3$ $mi{n}^{-1}$, $300mm$

To solve this question, we will use the following relation

$p_i=p_{infinite}$

$ln\left[\right(p_i-p_0\left)/\right(p_i-p_t\left)\right]=kt$ (1)

where $p_i$ is the pressure of the system at the infinite time (or a very long time), $p_0$ is the initial pressure, pt is the final pressure of the system, k is the rate constant and t is the time. Here the given equation is

$A\left(s\right)\to 2B\left(g\right)+C\left(g\right)$

As the starting material is a solid, therefore the initial pressure of the system will be zero. Hence the equation (1) will reduce to

$ln\left[\right(p_i\left)/\right(p_i-p_t\left)\right]=kt$ (2)

We are given that the pressure after a very long time is $225$ mm of $Hg$. So $p_i$ $=225$ mm. Also pressure after $20$ minutes is $150$ mm, $t=20$ minutes and $p_t$ $=p_{20}=150$ mm.

Substituting the above values in equation (2), we get

$ln\left[225/\right(225-150\left)\right]=k\times 20$

this gives $k$ $=0.05ln3$ per minute.

So the correct answer is $0.05ln3/min.$