In: Chemistry
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)?
The answer is either 9.9 atm or 1.1 atm. But my question is WHY IS ONE CHOICE INCORRECT since I get both in my calculations.
KC = 1.6
K = [CO][H2O] / [CO2][H2]
T = 1260
[CO2] = n/V = 0.038/1.5 = 0.0253 M
[H2] = 0.022/1.5 = 0.01466666 M
[CO] = 0
[H2O]
in equilibrium
[CO2] = 0.0253 - x
[H2] = 0.01466666 - x
[CO] = 0 + x
[H2O] = 0 + x
substitute in K
K = [CO][H2O] / [CO2][H2]
1.6 = (x*x)/((0.0253 - x)(0.01466666 - x))
solve for x
0.0253 *0.01466666 + (-0.0253 -0.01466666 )x + x^2 = 1/1.6 * x^2
0.000371-0.039966x 0.375x^2 = 0
x = 0.010273; 0.0963
note that we have 2 positive answers
one can't be since the substitution will give NEGATIVE concnetrations, which are impossible so...
for[CO2] = 0.0253 - x
option x1= 0.010273 --> 0.0253 -0.010273 = 0.015027 positive and possible
option x2 = 0.0963 --> 0.0253 -0.0963 = -0.071 negative and impossible
then
x1 is the only option
substitute in all equilibrium
[CO2] = 0.0253 - 0.010273 = 0.015027
[H2] = 0.01466666 - 0.010273 = 0.00439366
[CO] = 0 + 0.010273 = 0.010273
[H2O] = 0 + 0.010273 = 0.010273
for CO
[CO] = 0.010273 M
we can change M to atm as follows
P = M*RT
then
P = 0.010273*0.082*1260 = 1.06140636 atm or 1.1 which is the value you got...w e need to ignore 9.9 since this will be a false answer