Question

When $CO_{2(g)}$ is passed over red hot coke it partially gets reduced to $CO(g)$. Upon passing $0.5\, L $ of $CO_{2(g)}$ over red hot coke, the total volume of the gases increased to $700\, mL$. The composition of the gaseous mixture at $STP$ is

• A. $CO_2 = 300\, mL; CO=400 \,mL$
• B. $CO_2 = 0.0\, mL; CO=700 \,mL$
• C. $CO_2 = 200\, mL; CO=500 \,mL$
• D. $CO_2 = 350\, mL; CO=350 \,mL$

Answer 1

Answer: (A)

$CO_2+C \rightarrow 2CO$
Stoichoimetry ratio is $1 : 2$
AT $STP, P= 1 $ atm, $r= 273 K, R = 0.0821$
Initial moles of $CO_2. n$($CO_2$ initial) $=\frac{PV}{RT}$
$=\frac{1\times0.5}{0.0821\times273}=0.022$ mole
In final mixture no. of moles; $n$($CO_2/CO$ mixture)
$=\frac{1\times0.5}{0.0821\times273}=0.031$
Increase in volume is by $= 0.031 — 0.022$
$=0.009$ mole of gas
Final no. of moles of $CO$ i.e. $n_{\left(CO\,final\right)}$
$n_{\left(CO\,final\right)}=2n_{\left(CO_2\,initial\right)}-n_{\left(CO_2\,final\right)}$
$=2\left(0.002-n_{\left(CO_2\,final\right)}\right)\,...\left(i\right)$
$n_{\left(CO\,final\right)}=0.044-2n_{\left(CO_2\,final\right)}\,...\left(ii\right)$
$\therefore $ Now, $n_{\left(CO\,final\right)}+n_{\left(CO\,final\right)}=0.031$
$n_{\left(CO_2\,final\right)}=0.031-n_{\left(CO\,final\right)}\,...\left(iii\right)$
Substituting $\left(ii\right)$ in eq. $\left(i\right)$
$n_{\left(CO\,final\right)}=0.004-2\left[0.031-n_{\left(CO\,final\right)}\right]$
$n_{\left(CO\,final\right)}=0.044-0.062+2n_{\left(CO\,final\right)}$
$n_{\left(CO\,final\right)}=0.018\,mol.$
Volume of $CO=V=\frac{nRT}{P}=\frac{0.018\times0.0821\times273}{1}$
$=0.40\,Litre$
and volume of $CO_{2} = 0.7 litre - 0.4 \,litre$
$=0.3\,litre$
$\therefore CO_{2}=300\,mL, CO=400\,mL$