7.17. Kp =0.04 atm at 898 K for the equilibrium shown below. What is the equilibrium concentration ok C2H6 when it is placed in a flask at 4 atm pressure,and allowed to come to equilibrium.
C2H6(g) ⇌ C2H4(g) + H2(g).
7.17. Kp =0.04 atm at 898 K for the equilibrium shown below. What is the equilibrium concentration ok C2H6 when it is placed in a flask at 4 atm pressure,and allowed to come to equilibrium.
C2H6(g) ⇌ C2H4(g) + H2(g).
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1 Answer
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The equilibrium reaction is
C2?H6?(g) ? C2?H4?(g)+H2?(g).
Initial
4
0
0
Change
−x
x
x
Equilibrium
4−x
x
x
The expression for the equilibrium constant is Kp?= (?PC2?H4?) (PH2) / PC2?H6?.
Substituting the values in the above equation, we get
0.04=x2 / (4−x)? or x=0.38
Thus, the pressure of ethane is, PC2?H6?=3.62atm.
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