A) For the reaction: 3A(g)+2B(g)C(g) K c = 63.4 at a temperature of 377 C . Calc

Question

A) For the reaction: 3A(g)+2B(g)C(g) Kc = 63.4 at a temperature of 377 C .

Calculate the value of Kp. Express your answer numerically.

B) For the reaction X(g)+3Y(g)3Z(g) Kp = 2.91×102 at a temperature of 229 C .

Calculate the value of Kc. Express your answer numerically.

C) For the reaction 2CH4(g)C2H2(g)+3H2(g) K = 0.155 at 1774 C .

What is Kp for the reaction at this temperature? Express your answer numerically.

D) For the reaction N2(g)+3H2(g)2NH3(g) Kp = 4.40×103 at 271 C .

What is K for the reaction at this temperature? Enter your answer numerically.

E) The following reaction was performed in a sealed vessel at 772 C : H2(g)+I2(g)2HI(g)

Initially, only H2 and I2 were present at concentrations of [H2]=3.00M and [I2]=2.80M. The equilibrium concentration of I2 is 0.0400 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Express your answer numerically.

Thank you so much in advance!

Explanation / Answer

we know that

Kp = Kc (RT)^dn

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

A)

given

3A + 2B --> C

dn = 1 - 2 - 3 = -4

so

Kp = Kc (RT)^-4

so

Kp = 63.4 x (0.0821 x 650)^-4

Kp = 7.82 x 10-6

B)

given

X + 3Y ---> 3Z

dn = 3 - 3 - 1

dn = -1

so

Kp = Kc (RT)^-1

Kc = Kp (RT)

so

Kc = 2.91 x 10-2 x (0.0821 x 502)

Kc = 1.2

C)

given

2CH4 ---> C2H2 + 3H2

dn = 1 + 3 - 2 = 2

so

Kp = Kc (RT)^2

so

Kp = 0.155 x (0.0821 x 2047)^2

Kp = 4378

D)

given

N2 + 3H2 ---> 2NH3

dn = 2 - 1 - 3 = -2

so

Kp = Kc (RT)^-2

Kc = Kp (RT)^2

Kc = 4.4 x 10-3 ( 0.0821 x 544)^2

Kc = 8.77

E)

H2 + I2 --> 2HI

using ICE table

at equilibrium

[H2] = 3 - x

[I2] = 2.8 - x

[HI] = 2x

given

[I2] = 0.04

so

2.8 - x = 0.04

x = 2.76

so

[H2] = 3 - x = 3 - 2.76 = 0.24

[I2] = 2.8 - x = 0.04

[HI] = 2x = 2 * 2.76 = 5.52

now

Kc = [HI]^2 / [H2] [I2]

Kc = [5.52]^2 / [0.24] [0.04]

Kc = 3174

so

the value of equilibrium constant is 3174

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