What the correct expression for the equilibrium constant (K_c) for the reaction

Question

What the correct expression for the equilibrium constant (K_c) for the reaction between carbon and hydrogen to form methane shown here? C(s) + 2 H_2(*g) CH_4(g) a) K_c = [CH_4]/[H_2] b) K_c = [CH_4]/[C][H_2] c) K_c = [CH_4]/[C][H_2]^2 d) K_c = [CH_4]/[H_2]^2 The equilibrium constant for the reaction shown here is K_c = 1.0 times 10^3. A reaction mixture at equilibrium contains [A] = 1.0 times 10^-3 M. What is the concentration of B in the mixture? A(g) B(g) a) 1.0 times 10^-3 M b) 1.0 M c) 2.0 M d) 1.0 times 10^3 M Use the data below to find the equilibrium constant (K_c) for the reaction A(g) 2 B(g) + C(g). A(g) 2 X(g) + C(g) K_c = 1.55 B(g) X(g) K_c = 25.2 a) 984 b) 26.8 c) 6.10 times 10^-4 d) 2.44 times 10^-3 The reaction shown here has a K_p = 4.5 times 10^2 at 825 K. Find K_c for the reaction at this temperature. CH_4(g) + CO_2(g) 2 CO(g) + 2 H_2(g) a) 0.098 b) 2.1 times 10^6 c) 6.6 d) 4.5 times 10^-2 Consider the reaction between NO and CI_2 to form NOCI: 2 NO(g) + CI_2(g) 2 NOCI(g) A reaction mixture at a certain temperature initially contains only [NO] = 0.50 M and [CI_2] = 0.50 M. After the reaction comes to equilibrium, the concentration of NOCI is 0.30 M. Find the value of the equilibrium constant (K_c) at this temperature. a) 11 b) 4.3 c) 6.4 d) 0.22 For the reaction 2 A(g) B(g), the equilibrium constant is K_p = 0.76. A reaction mixture initially contains 2.0 atm of each gas (P_A = 2.0 atm and P_B = 2.0 atm). Which statement is true of the reaction mixture? a) The reaction mixture is at equilibrium. b) The reaction mixture will proceed toward products. c) The reaction mixture will proceed toward reactants. d) It is not possible to determine from the information given the future direction of the reaction mixture. Consider the reaction between iodine gas and chlorine gas to form iodine monochloride: I_2(g) + CI_2(g) 2 ICI(g) K_p = 81.9 (at 298 K) A reaction mixture at 298 K initially certains P_I2 = 0.25 atm and P_CI2 = 0.25 atm. What is the partial pressure of iodine monochloride when the reaction reaches equilibrium? a) 0.17 atm b) 0.64 atm c) 0.41 atm d) 2.3 atm

Explanation / Answer

1)

C(s) + 2H2 ---> CH4

we know that

solids are not considered for the expression of Kc

so

Kc = [CH4] / [H2]^2

the answer is option d ) Kc = [CH4] / [H2]^2

2)

A ---> B

Kc = [B] / [A]

using given values

1 x 10^3 = [B] / 10-3

[B] = 1 M

so

the answer is b) 1.0 M

3)

A ---> 2 X + C

K1 = [X]^2 [C] / [A]

B ---> X

K2 = [X] / [B]

A ---> 2B + C

K3 = [B]^2 [C] / [A]

we can see that

K3 = K1 / ( K2)^2

using given values

K3 = 1.55 / ( 25.2)^2

K3 = 2.44 x 10-3

so

the answer is d) 2.44 x 10-3

4)

we know that

Kp = Kc (RT)^dn

here

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

given

CH4 + C02 ---> 2 CO + 2H2

in this case

dn = 2 + 2 - 1 - 1

dn = 2

so

Kp = Kc (RT)^dn

4.5 x 10^2 = Kc ( 0.0821 x 825)^2

Kc = 0.098

so

the answer is a) 0.098

5)

2 NO + Cl2 --> 2 NOCl

using ICE table

initial conc of NO , Cl2 , NOCl are 0.5 , 0.5 , 0

change in conc of NO ,Cl2 , NOCl are -2x , -x , + 2x

equilibrium conc of NO , Cl2 , NOCl are 0.5-2x , 0.5-x , 2x

given

[NOCl]eq = 0.3

so

2x = 0.3

x = 0.15

now

[NO]eq = 0.5-2x = 0.5-0.3 = 0.2

[Cl2]eq = 0.5 -x = 0.5- 0.15 = 0.35

now

Kc = [NOCl]^2 / [NO]^2 [Cl2]

Kc = [0.3]^2 / [0.2]^2 [0.35]

Kc = 6.428

so

the answer is c) 6.4

6)

2A ---> B

the reaction quotient is given by

Q = [pB] / [pA]^2

initially

Q = [2]/ [2]^2

Q = 0.5

now

we know that

if Q < Kp , then the reaction mixture will proceed toowards the products

if Q = Kp , then the reaction mixture is at equilibrium

if Q > Kp , then the reaction mixture will proceed towards the reactants

in this vase

Q = 0.5 < Kp = 0.76

so

the answer is b) the reaction mixture will proceed towards the products

7)

I2 + Cl2 ---> 2 ICl

using ICE table

at equilibrium

pICl = 2x

pI2 = 0.25 - x

pCl2 = 0.25 - x

now

Kp = [pICl]^2 / [pI2] [pCl2]

81.9 = [2x]^2 / [0.25-x] [0.25-x]

81.9 = ([ 2x/ (0.25-x) ]^2

9.05 = 2x / (0.25-x)

2x = 9.05 (0.25-x)

x = 0.205

now

at equilibrium

pICl = 2x = 2 * 0.205 = 0.41 atm

so

the answer is C) 0.41 atm

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