A mixture with [N_2O_4] = 0.90 and [N_2O| = 0.134 it at equilibrium as follow: N

Question

A mixture with [N_2O_4] = 0.90 and [N_2O| = 0.134 it at equilibrium as follow: N_2O_4(g) 2No_2(g) What is the value of K_c for the reaction? The value of K_p for the following equilibrium is 1.1 Times 10^-1 at 600 degree C: 2 SO_3(g) 2 SO_2(g) + O_2(g) What is the value of K_c foe the reaction? (R = 0.0S206 Latm/mol K) 8 K_p for the following reaction equilibrium at 425.4 degree C is 54.5. H_2 (g) + I_2 (g)g 2HI(g)G If the partial pressures in a mixture was found to be pH_2 = 0.75 atm, pI_2 = 0.75 atm, and pHI = 2.5 atm, determine whether the mixture is at equilibrium or not. If the mixture is not at equilibrium, tell the direction of the shill, and calculate the equilibrium partial pressures of H_2, I_2, and HI. The value of K_p is 0.113 at 25 degree C for the following equilibrium If the total pressure is = 8.0 atm. what are the partial pressures of N_2O_4 and NO_2? Predict the effect of each of the following changes on the equilibrium increasing pressure by decreasing volume at constant temperature decreasing pressure by increasing volume at constant temperature removing some CO from the mixture increasing temperature at constant pressure adding a catalyst.

Explanation / Answer

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9.
Let partial pressure of NO2 be x atm
Then partial pressure of N2O4 will be 8-x atm since total pressure is 8 atm

Kp = p(NO2)^2 / p(N2O4)
0.113 = x^2 / (8-x)
x^2 =0.904 - 0.113 x
x^2 + 0.113x - 0.904 = 0

Solving the above quadratic equation, the positive value of x is:
x = 0.896

So,
partial pressure of NO2 is 0.896 atm
partial pressure of N2O4 = 8-x = 8 - 0.896 = 7.104 atm

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