Question

Calculate the change in pressure (in atm) when 2 mole of NO and 16 g $O_2$, in a 6.25 litre contain Originally at ${27}^{o}C$ react to produce the maximum quantity of NO, possible according to the equation
$2NO\left(g\right)O_2,\left(g\right)\to 2N{O}_2\left(g\right)$ (Take $R=\frac{1}{12}$ It atm/mot K)

Answer 1

$2NO+O_2⟶2N{O}_2$

Initially, Moles of $NO=2$ $moles$

Moles of $O_2=\frac{16g}{32g/mol}=0.5$ $mole$

Total moles $=2+0.5=2.5$ $moles$

$\therefore P_i=\frac{n_{i}RT}{V_i}=2.5\times \frac{1}{12}\times \frac{(27+273)}{6.25}$

$P_i=10$ $atm$

$2NO+O_2⟶2N{O}_2$ Initial moles consumed/ produced moles finally:

$2$ $moles$ $0.5$ $moles$ $0$ $mole$

$1$ $mole$ $0.5$ $mole$ $1$ $mole$

$(2-1)$ $(0.5-0.5)$ $(0+1)$

$1$ $mole$ $0$ $mole$ $1$ $mole$

Total moles after reaction $=2$ $moles$

$P_f=\frac{n_{f}RT}{V_f}=2\times \frac{1}{12}\times \frac{(27+273)}{6.25}=8$ $atm$

Change in pressure $=\Delta P=P_f-P_i=8-10=-2$ $atm$

$-ve$ sign indicates decrease in pressure.