The equilibrium constant, Kc, for the reaction: C(s) + CO_2 (g) rlhar 2 CO (g) i

Question

The equilibrium constant, Kc, for the reaction: C(s) + CO_2 (g) rlhar 2 CO (g) is 0.113 at 1100 K. What is the concentration of CO in a mixture at equilibrium if the CO_2 concentration is 0.035 M? If K_c = 7.5 times 10^-9 at 1000 K for the reaction: N_2(g) + O_2(g) rlhar 2 NO(g) What is K_p at 1000 K for the reaction, and; What is K_c for the reaction? 1/2 N_2(g) + O_2(g) rlhar NO(g) at 1000 K? The equilibrium constant, K_c, for the reaction H_2(g) + I_2(g) rlhar 2 HI(g) is 55 at 425 degree C. If a 0.40 mol sample of HI was introduced into a 1.00 mL reaction vessel at 425 degree C, what are the equibrium concentrations of H_2, I_2 and HI?

Explanation / Answer

Q7.

K = CO^2 / (CO2)

K = 0.113

find CO if...

[CO2] = 0.035

K = CO^2 / (CO2)

0.113= (CO)^2 / (0.035)

CO = sqrt(0.113/0.035)

[CO] = 1.7968225 M

Q8.

K = 7.5*10^-9

K = [NO]^2 / [N2][O2]

find Kp:

Kp = Kc*(RT)^dn

dn = moles of product - mol reactants = 2-(1+1) = 0

so

Kp = Kc = 7.5*10^-9

then

b)

for

half equation

simply....

Knew = K1^(1/2) = (7.5*10^-9)^0.5 = 0.00008660254

Q9

K = [HI]^2 / [H2][I2]

K = 55

[HI] = 0.4 /(1*10^-3) = 400M

find concentrations

so.. initially

[HI] = 400 M

[I2] = 0

[H2] = 0

in equliibrium

[HI] = 400 - 2x

[I2] = 0 + x

[H2] = 0 + x

substitute

K = [HI]^2 / [H2][I2]

55 = (400-2x)^2/ (x*x)

sqrt(55) = (400-2x) / x

7.4161x = 400 - 2x

9.4161x = 400

x = 400/9.416 = 42.48088

so

[HI] = 400 - 2*42.48088 = 315.03824

[I2] = 0 + x = 42.48088

[H2] = 0 + x = 42.48088

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