Question

The decomposition of $C{l}_2{O}_7$ at $400$K in the gas phase to $C{l}_2$ and $O_2$ is a first order reaction. (i) after $55$ seconds at $400$K, the pressure of $C{l}_2{O}_7$ falls from $0.062$ atm to $0.044$ atm. Calculate the rate constant. (ii) Calculate the pressure of $C{l}_2{O}_7$, after $100$ sec. of decomposition at the temperature.

Answer 1

The given reaction is:-

$2C{l}_2{O}_7\left(g\right)⟶2C{l}_2\left(g\right)+7O_2\left(g\right)$

Given that the reaction is of first order, so rate law is given by,

Rate=$K\left[C{l}_2{O}_7\right]$;$K$ is rate constant

Now, given that after $55s$ pressure of $C{l}_2{O}_7$ is $0.044$ atm and at $t=0$ ; the pressure of Cl_2O_7$was$ 0.062$ atm.

For first order gaseous reaction, we have,

$K=\frac{1}{t}ln\frac{P_0}{P_t}$

where, $K$= rate constant; $t$=times

$P_0$= pressure of reactant gas at $t=0$

$P_t$ pressure of reactant gas at $t=t$

So, $K=\frac{1}{55}ln\left(\frac{0.062}{0.44}\right)$

$\Rightarrow K=\frac{1}{55}\times 0.343=0.0062s^{-1}$

Now, after $t=100s$, we have,

$K=\frac{1}{t}ln\left(\frac{P_0}{P_t}\right)$

$\Rightarrow 0.0062=\frac{1}{100}ln\left(\frac{0.062}{P_t}\right)$

$\Rightarrow 0.62=ln\left(\frac{0.062}{P_t}\right)$

taking antilog on both sides:-

$\Rightarrow \frac{0.062}{P_t}=e^{0.62}$

$\Rightarrow P_t=\frac{0.062}{e^{0.62}}=\frac{0.062}{1.86}=0.033$ atm.