Consider the reaction: 2NO_2 (g) Equilibrium N_2 O_4 (g) If an equilibrium mixtu

Question

Consider the reaction: 2NO_2 (g) Equilibrium N_2 O_4 (g) If an equilibrium mixture was found to have [NO_2] = 0.0125M and [N_2 O_4] = 0.0337 M, what is Kc for the reaction The equilibrium constant, K_c, for the reaction: C(s) + CO_2 (g) Equilibrium 2 CO (g) is 0.113 at 100 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) Equilibrium 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 + 1/2 O_2 (g) Equilibrium NO (g) at 1000 K? The equilibrium constant, K_c, for the reaction

Explanation / Answer

Q6

Kc = [products]/[reactnats]

Kc = [N2O4] / [NO2]^2

Kc = (0.0337)/(0.0125^2)= 215.68

Q7

Kc = [CO]^2 / [CO2]

ignore solids, since activity = 1

so..

0.113 = [CO]^2 / (0.035)

[CO] = sqrt(0.113 *0.035)

[CO] = 0.0628 M

Q8

Kp = Kc*(RT)^dn

dn = 2-(1+1) = 2-2 = 0

Kp = Kc = 7.5*10^-9

b)

Kc for reaction divide dby 2... means sqare

so

Knew = sqrt(Kc) = sqrt(7.5*10^-9) = 0.00008660254

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