Carbon monoxide in water gas reacts with steam: (g) + (g) β (g) + (g); Keq = 0.3333. Two volumes of β Chemical Equilibrium Chemistry Question
Question
Carbon monoxide in water gas reacts with steam: $CO$(g) + $H_2O$(g) β $CO_2$(g) + $H_2$(g); Keq = 0.3333. Two volumes of water gas (1:1 $CO$:$H_2$) are mixed with five volumes of steam. The volume ratio of the four gases at equilibrium is:
π‘ Solution & Explanation
Step 1 - Set up the ICE table For \(\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}\): Initial volumes: CO = 1, H2O = 5, CO2 = 0, H2 = 1 (from 2 vol water gas + 5 vol steam) Let x = volumes of CO that react: | | CO | H2O | CO2 | H2 | |--|--|--|--|--| | Initial | 1 | 5 | 0 | 1 | | Change | -x | -x | +x | +x | | Equilibrium | 1-x | 5-x | x | 1+x | Step 2 - Write the Keq expression Since Ξn_gas = 0, volume terms cancel: \[K_{\text{eq}} = \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]} = \frac{x(1+x)}{(1-x)(5-x)} = \frac{1}{3}\] Step 3 - Solve the quadratic Cross-multiplying: \[3x(1+x) = (1-x)(5-x)\] \[3x + 3x^2 = 5 - 6x + x^2\] \[2x^2 + 9x - 5 = 0\] \[(2x - 1)(x + 5) = 0\] \[x = 0.5 \quad (\text{reject } x = -5)\] Step 4 - Find equilibrium volumes \[\ce{CO} : \ce{H2O} : \ce{CO2} : \ce{H2} = 0.5 : 4.5 : 0.5 : 1.5 = \boxed{1:9:1:3}\] Step 5 - Explain all options * **Option (A) 1:1:1:1**: Incorrect. Requires x = 0 (no reaction) and equal initial amounts, which doesn't match. * **Option (B) 1:3:1:1**: Incorrect. Incorrect solution to the quadratic equation. * **Option (C) 1:15:1:5**: Incorrect. Mathematical calculation error. * **Option (D) 1:9:1:3**: Correct. Solving the quadratic from Keq = 1/3 gives x = 0.5, yielding this ratio.