When CO2 (g) is passed over red hot coke it partially gets reduced to CO(g). Upon passing 0.5 L of CO2 (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) CO2 = 300 mL; CO = 400 mL
(b) CO2 = 0.0 mL; CO = 700 mL
(c) CO2 = 200 mL; CO = 500 mL
(d) CO2 = 350 mL; CO = 350 mL
Correct answer: (a) CO2 = 300 mL; CO = 400 mL
Explanation:
CO2 + C → 2CO
Stoichoimetry ratio is 1 : 2
AT STP, P = 1 atm, T = 273 K, R = 0.0821
Initial moles of CO2; n(CO2 initial) = \(\frac{PV}{RT}\)
= \(\frac{1 \times 0.5}{0.0821 \times 273}\)
= 0.022 mole
In final mixture no. of moles; n(CO2 /CO mixture)
= \(\frac{1 \times 0.7}{0.0821 \times 273}\)
= 0.031 mole
Increase in volume is by = 0.031 – 0.022
= 0.009 mole of gas
Final no. of moles of CO i.e. n(CO final)
n(CO final) = 2n(CO2 initial) – n(CO2 final)
= 2(0.022 - n(CO2 final)) ......(i)
n(CO final) = 0.044 - 2n(CO2 initial) ......(ii)
∴ Now, n(CO final) + n(CO final) = 0.031
n(CO2 final) = 0.031 - n(CO final) .....(iii)
Substituting (ii) in eq. (i)
n(CO final) = 0.044 - 2[0.031 - n(CO final)]
n(CO final) = 0.044 - 0.062 + n(CO final)
n(CO final) = 0.018 mol.
Volume of CO = V = \(\frac{nRT}{P}\)
= \(\frac{0.018 \times 0.0821 \times 273}{1}\)
= 0.40 Litre
and volume of CO2 = 0.7 litre - 0.4 litre
= 0.3 litre
∴ CO2 = 300 mL, CO = 400 mL