Question

part a.)

For the reaction

2CH4(g)⇌C2H2(g)+3H2(g)

Kc=0.140 mol2 L−2 at 1517 ∘C . What is Kp for the reaction at this temperature?

Enter your answer numerically.

part b.)

What is the unitless thermodynamic equilibrium constant K for the reaction in part A?

Express your answer numerically.

part c.)

For the reaction

N2(g)+3H2(g)⇌2NH3(g)

Kp=5.70×10−3 bar−2 at 316 ∘C . What is Kc for the reaction at this temperature?

Enter your answer numerically.

part d.)

What is the unitless thermodynamic equilibrium constant K for the reaction in part C?

Express your answer numerically.

Answer 1

a)

To change Kc into Kp

Kp = Kc (RT)^Δn

Kp = equilibrium constant in terms of partial pressures (atm)
Kc = equilibrium constant in terms of concentrations(molarity), 0.140 mol2 L−2
R = ideal gas constant ( 0.082 atm.L / K.mol)
T = absolute temperature, 1517 C (1790 K)
Δn = Σ n products - Σ n reactants
Δn = (4) - (2) = 2 (the solid is not taken into account)

Kp = Kc (RT)^Δn
Kp = (0.140 mol2 L−2) {(0.082 atm.L / K.mol)(1790 K}^2

= 3016 atm^2

B) unitless thermodynamic equilibrium constant K for the reaction in part A is 3016

C)

N2(g)+3H2(g)⇌2NH3(g)

Kp=5.70×10−3 bar−2 at 316 ∘C .

We know that Kp = Kc * (RT) ^Δn

Where T = Temperature = 316 o C = 316 + 273 =589 K

R = gas constant

8.3144598(48)×10−2

L⋅bar⋅K−1⋅mol−1

Δn = No . of moles of gaseous products - No . of moles ofgaseous reactants

     = 2 - ( 1+3)

     =-2

Kp = Kc (RT)^Δn
5.70×10−3 bar−2 = Kc {(0.08314 bar .L / K.mol)(589 K}^-2

5.70×10−3 bar−2 / 2398 (bar .L / .mol)^-2 = Kc

5.70×10−3 /0.000417 L^2 /.mol^2= Kc

= 13.66 L^2 /.mol^2

D) unitless thermodynamic equilibrium constant K for the reaction in part A is13.66