Question

Consider the following equilibrium: CO2(g) + H2(g) CO(g) + H2O(g); Kc = 1.6 at 1260 K Suppose 0.038 mol CO2 and 0.022 mol H2 are placed in a 1.50-L vessel at 1260 K. What is the equilibrium partial pressure of CO(g)? (R = 0.0821 L · atm/(K · mol)). The answer is either 9.9 atm or 1.1 atm when I solve using the quadratic equation but I can't decide which answer is correct. Please explain how you eliminated the incorrect answer. thank you...

Answer 1

CO2(g) + H2(g) ----------> CO(g) + H2O(g) Kc = 1.6

Intial 0.038 mol/1.5 L 0.022 mol/1.5 L 0 0

= 0.0253 M = 0.0146 M

at equilibrium 0.0253 -x 0.0146 -x x x

Kc = x.x / ( 0.0253 -x ) ( 0.0146 -x)

1.6 = x² / (x² - 0.04 x + 0.00037)

1.6x² - 0.064 x + 0.0006 =  x²

0.6x² - 0.064 x + 0.0006 = 0

On solving, x = 0.0102 M

Therefore,

equilibrium concentration of CO (g) =  0.0102 M

PV = nRT

P = (n/V) RT

= MRT

= 0.0102 M x 0.0821 L.atm/mol/K x 1260 K

= 1.1 atm

P = 1.1 atm

Therefore, equilibrium partial pressure of CO(g) = 1.1 atm