Question

Consider the equilibrium C2H6(g) ↔ C2H4(g) + H2(g) . At 1000K and a constant total pressure of 1 bar, H2(g) is introduced into the reaction vessel. The total pressure is held constant at 1 bar and at equilibrium the composition of the mixture in mole percent is H2 : 26% ; C2H4: 26% ; C2H6 : 48%Calclate KP at 1000 K.

Answer 1

`K_(p) (1000K) = 0.141`

Step-by-step explanation:

From the reaction:

C₂H₆(g) ⇆ C₂H₄(g) + H₂(g)

48% 26% 26%

Knowing the composition of the mixture at equilibrium (at 1000K), we can calculate the equilibrium constant in terms of mole fraction:

`K_(x) = \frac{X_{C_(2)H_(4)} \cdot X_{H_(2)}}{X_{C_(2)H_(6)}}`

where X: mole fraction of C₂H₆(g), C₂H₄(g) and H₂(g)

`K_(x) = (0.26 \cdot 0.26)/(0.48)`

`K_(x)= 0.141`

Now, the equilibrium constant in terms of pressure can be calculated using the equilibrium constant in terms of mole fraction:

`K_(p) = K_(x) \cdot (P_(T)) ^ {\Delta n}`

where
`P_(T)`: total pressure and Δn: number of gaseous moles of product - number of gaseous moles of reactant

`K_(p) = 0.141 \cdot (1) ^ {2-1}`

`K_(p) (1000K) = 0.141`

Have a nice day!