HW14_#7 PLEASE SHOW ALL WORK The equilibrium constant, Kc, is calculated using m

Question

HW14_#7 PLEASE SHOW ALL WORK

The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, Kp, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation Kp = Kc(RT)^Delta n 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 N2(g) + 3H_2(g) 2NH_3(g) for which Delta n = 2 - (1 + 3) = -2. For the reaction 3A(g) + 2B(g) C(g) Kc = 40.2 at a temperature of 123 degree C . Calculate the value of Kp. Express your answer numerically. For the reaction X(g) + 3Y(g) 2Z(g) Kp = 2.92 Times 10^-2 at a temperature of 305 degree C . Calculate the value of Kc. Express your answer numerically.

Explanation / Answer

A) given reaction is

3A (g) + 2B (g) ---> C (g)

now

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

so

dn = 1 - 2 - 3

dn = -4

now

given

T = 123 C

T = 123 + 273 K

T = 396 K

Kp = Kc (RT)^dn

Kp = 40.2 ( 0.08206 x 396)^-4

Kp = 3.6 x 10-5

so

the value of Kp is 3.6 x 10-5

B)

given reaction is

X (g) + 3Y (g) ---> 2Z (g)

now

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

so

dn = 2 - 1 -3

dn = -2

now

Kp = Kc (RT)^dn

Kp = Kc (RT)^-2

Kc = Kp (RT)^2

now

given

Kp = 2.92 x 10-2

R = 0.08206

T = 305 C

T = 305 + 273

T = 578 K

now

Kc = Kp (RT)^2

Kc = 2.92 x 10-2 ( 0.08206 x 578)^2

Kc = 65.69

so

the value of Kc is 65.69

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