concentration: how do I solve both of these The equilibrium constant, K_ c, is c

Question

concentration: how do I solve both of these

The equilibrium constant, K_ c, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, K_ p, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation where R = 0.08206 L.atm/(K.mol), T is the absolute temperature, and An is the change in the number of moles of gas (sum moles products - sum moles reactants). For example, consider the reaction N_2 (g) + 3H_2 (g) equivalent 2NH_3 (g) For the reaction X(g) + 3Y(g) equivalent 2Z(g) K_ p = 3.92x10"2 at a temperature of 103 degree C. Calculate the value of K_ c. Express your answer numerically.

Explanation / Answer

1)

3A + 2B -->> C

we know that

dn = moles of gases in products - moles of gases in reactants

so

dn = 1 - 2 - 3

dn = -4

now

Kp = Kc (RT)^dn

given

T = 91 C

T = 91 + 273

T = 364 kelvin

so

Kp = 69 x ( 0.0821 x 364)^-4

Kp = 8.651 x 10-5

so

the value of Kp is 8.651 x 10-5

2)

X + 3Y --->> 2Z

we know that

dn = moles of gases in products - moles of gases in reactants

so

dn = 2 - 1 - 3

dn = -2

now

Kp = Kc (RT)^dn

given

T = 103 C

T = 103 + 273

T = 376 kelvin

so

Kp = Kc (RT)^-2

Kc = Kp (RT)^2

Kc = 3.92 x 10-2 x ( 0.0821 x 376)^2

Kc = 37.355

so

the value of Kc is 37.355

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