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

Answer:

$K_p= 3966.01$

Explanation:

The relation between Kp and Kc is given below:

$K_p= K_c\times (RT)^{\Delta n}$

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:

$K_p= 0.140\times (0.082057\times 2051.15)^{2}$

$K_p= 3966.01$