Question

The reaction 2NO(g)+Cl2(g)→2NOCl(g) is carried out in a closed vessel. If the partial pressure of NO is decreasing at the rate of 21 torr/min , what is the rate of change of the total pressure of the vessel

Answer 1

Answer : The rate of change of the total pressure of the vessel is, 10.5 torr/min.

Explanation : Given,

$\frac{d[NO]}{dt}$ =21 torr/min

The balanced chemical reaction is,

$2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)$

The rate of disappearance of $NO$ = $-\frac{1}{2}\frac{d[NO]}{dt}$

The rate of disappearance of $Cl_2$ = $-\frac{d[Cl_2]}{dt}$

The rate of formation of $NOCl$ = $\frac{1}{2}\frac{d[NOCl]}{dt}$

As we know that,

$\frac{d[NO]}{dt}$ =21 torr/min

So,

$-\frac{d[Cl_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}$

$\frac{d[Cl_2]}{dt}=\frac{1}{2}\times 21torr/min=10.5torr/min$

And,

$\frac{1}{2}\frac{d[NOCl]}{dt}=\frac{1}{2}\frac{d[NO]}$

$\frac{d[NOCl]}{dt}=\frac{d[NO]}=21torr/min$

Now we have to calculate the rate change.

Rate change = Reactant rate - Product rate

Rate change = (21 + 10.5) - 21 = 10.5 torr/min

Therefore, the rate of change of the total pressure of the vessel is, 10.5 torr/min.