Question

when 1.50 mol CO2 and 1.50 mol H2 are placed in a 3.00-L container at 395°C, the following reaction occurs: CO2(g) + H2(g) = CO(g) + H2O(g). If Kc = 0.802, what are the concentrations of each substance in the equilibrium mixture?

Answer 1

C_M (CO₂)= n /V = 1,5 / 3 = 0,5 mol/l

C_M (H₂)= n /V = 1,5 / 3 = 0,5 mol/l

CO2 + H₂ = CO + H₂O

Concentration for equilibrium:

0,5-x — 0,5-x — x — x

$K_c = \frac {[CO][H_2O]}{[CO_2][H_2]}$

$0,802= \frac {[0,5][0,5]}{[0,5-x][0,5-x]}$

0,802 * (0,25 - 0,5x + x²) = 0,25

0,802x² - 0,401x + 0,2005 = 0,25

0,802x² - 0,401x - 0,0495 = 0

$X_{1,2} = \frac{-b ±\sqrt{b² -4ac} }{2a}$

$X =\frac{0,401± \sqrt{(-0,401)²-4*0,802*0,0495}}{2*0,802}$

$X= 0,278$

ANSWER:

C_M (CO₂) = 0,5 - 0,278 = 0,222 mol/l

C_M (H₂) = 0,5 - 0,278 = 0,222 mol/l

C_M (CO) = 0,278 mol/l

C_M (H₂O) = 0,278 mol/l