the value of K_c for the reaction? The value of K_p for the following equilibriu

Question

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 = 0.08206 Latm/mol K) 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 pm = 2.5 atm, determine whether the mixture is at equilibrium or not. If the mixture is not at equilibrium, tell the direction of the shift, 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 N_2O_4(g) 2NO_2(g) If the total pressure is = 8.0 atm, what are the partial pressures of N_2 O_4 and NO_2? Predict the effect of each of the following changes on the equilibrium CH_4(g)+ H_2O(g) 3H_2(g)g + CO(g), Delta H degree = + 203kJ 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. For the equilibrium in part (a), each of the following changes will increase the equilibrium partial passes of HBr. Choose the change that will cause the gases increase in the pressure of HBr, and explain in choice.

Explanation / Answer

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7.
Use:
Kp= Kc*(RT)^delta n

delta n is difference in number of gaseous molecules in product and reactant
delta n= 2+1 -2 = 1

T = 600 oC = (600+273) K= 873 K

Kp= Kc*(RT)^delta n
1.1*10^-1 = Kc* (0.08206*873)^1
Kc = 1.1*10^-1/ (0.08206*873)
Kc = 1.54*10^-3
Answer: 1.54*10^-3

8.
Qp = p(HI)^2 / {p(H2)*p(I2)}
       = (2.5)^2 / {(0.75)*(0.75)}
       = 11.11

Since Qp is less than Kp, reaction will shift towards right

H2         +     I2 ---> 2HI
0.75             0.75          2.5   (initial)
0.75-x        0.75-x        2.5+x (at equilibrium)

Kp= p(HI)^2 / {p(H2)*p(I2)}
54.5 = (2.5+x)^2 / (0.75-x)^2
7.38 = (2.5+x) / (0.75-x)
5.54 - 7.38 x = 2.5 + x
x= 0.36

So, equilibrium partial pressures are:
p(HI) = 2.5+x = 2.5+0.36 = 2.86 atm
pH2 = 0.75-x = 0.75-0.36= 0.39 atm
pI2 = 0.75-x = 0.75-0.36= 0.39 atm

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