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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.

Respuesta :

whitneytr12

Answer:

[tex]K_{p} (1000K) = 0.141 [/tex]

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:          

[tex] K_{x} = \frac{X_{C_{2}H_{4}} \cdot X_{H_{2}}}{X_{C_{2}H_{6}}} [/tex]

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

[tex] K_{x} = \frac{0.26 \cdot 0.26}{0.48} [/tex]

[tex] K_{x}= 0.141 [/tex]                      

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

[tex] K_{p} = K_{x} \cdot (P_{T}) ^ {\Delta n} [/tex]  

where [tex]P_{T} [/tex]: total pressure and Δn: number of gaseous moles of product - number of gaseous moles of reactant  

[tex] K_{p} = 0.141 \cdot (1) ^ {2-1} [/tex]        

[tex] K_{p} (1000K) = 0.141 [/tex]

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