Question

For chemical reactions involving ideal gases, the equilibrium constant K can be expressed either in terms of the concentrations of the gases (in M) or as a function of the partial pressures of the gases (in atmospheres). In the latter case, the equilibrium constant is denoted as Kp to distinguish it from the concentration-based equilibrium constant Kc (sometimes referenced as just K).For the reaction 2CH4(g)⇌C2H2(g)+3H2(g) Kc = 0.140 at 1778 ∘C . What is Kp for the reaction at this temperature? Express your answer numerically.

Answer 1

`K_p= 3966.01`

Step-by-step explanation:

The relation between Kp and Kc is given below:

`K_p= K_c* (RT)^(\Delta n)`

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

`2CH_4_((g))\rightleftharpoons C_2H_2_((g))+3H_2_((g))`

Given: Kc = 0.140

Temperature = 1778 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (1778 + 273.15) K = 2051.15 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (3+1)-(2) = 2

Thus, Kp is:

`K_p= 0.140* (0.082057* 2051.15)^(2)`

`K_p= 3966.01`