The equilibrium constant for (g) + (g) ⇌ (g) + (g) is 3.0 at 500 K. In a 2.0 L vessel, 60 g of water — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Determine initial moles of each species Water gas (equimolar \ce{CO}:\ce{H2}) has average molar mass = (28+2)/2 = 15 g/mol. 60 g water gas = 60/15 = 4 mol total → 2 mol \ce{CO} + 2 mol \ce{H2}. 90 g steam: 90/18 = 5 mol \ce{H2O}. Step 2 - Set up ICE table (in moles, vessel V = 2.0 L) Reaction: \[\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}\] \[\begin{array}{lcccc} & \ce{CO} & \ce{H2O} & \ce{CO2} & \ce{H2} \ \hline \text{Initial (mol)} & 2 & 5 & 0 & 2 \ \text{Change (mol)} & -x & -x & +x & +x \ \text{Equilibrium (mol)} & 2-x & 5-x & x & 2+x \ \hline \end{array}\] Step 3 - Apply Kc expression and solve for x Since Δng = 0, concentrations and mole ratios give the same Kc: \[K_c = \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]} = \frac{x(2+x)}{(2-x)(5-x)} = 3\] Expanding: \[x(2+x) = 3(2-x)(5-x) = 3(10 - 7x + x^2)\] \[2x + x^2 = 30 - 21x + 3x^2\] \[2x^2 - 23x + 30 = 0\] Using the quadratic formula: \[x = \frac{23 \pm \sqrt{529 - 240}}{4} = \frac{23 \pm \sqrt{289}}{4} = \frac{23 \pm 17}{4}\] \(x = 10\) (rejected: exceeds initial moles) or \(\boxed{x = 1.5\text{ mol}}\) Step 4 - Calculate equilibrium concentration of H2 \[[\ce{H2}]_\text{eq} = \frac{2 + x}{V} = \frac{2 + 1.5}{2.0} = \frac{3.5}{2.0} = \boxed{1.75\text{ M}}\] Step 5 - Evaluate all options - **Option (A) 1.75 M**: Correct. x = 1.5 mol, [H2] = 3.5/2.0 = 1.75 M. - **Option (B) 3.5 M**: Incorrect. This is total moles of H2 (3.5), not concentration. - **Option (C) 1.5 M**: Incorrect. This is the value of x (mol reacted), not [H2]. - **Option (D) 0.75 M**: Incorrect. Computation error (x = 0.75 holds for [CO] initial = 1 M setup).