Asked by Bob
The equilibrium constant, Kp, equals 3.40 for the isomerization reaction:
cis-2-butene trans-2-butene.
If a flask initially contains 0.250 atm of cis-2-butene and 0.145 atm of trans-2-butene, what is the equilibrium pressure of each gas?
cis-2-butene trans-2-butene.
If a flask initially contains 0.250 atm of cis-2-butene and 0.145 atm of trans-2-butene, what is the equilibrium pressure of each gas?
Answers
Answered by
DrBob222
.....cis-butene==> trans-butene
I......0.250........0.145
C.......-x...........+x
E......0.250-x.....0.145+x
Substitute the E line into Kp expression and solve for x, 0.250-x and 0.145+x.
How did I know to make the C line -x......+x an not +x......-x; i.e., how do I know it will shift from cis to trans and NOT trans to cis.
I calculated Qp = 0.145/0.250 = 0.58 and compared that value with Kp. It is smaller than Kp and that tells me the numerator is too small and the denominator is too large so I know the reaction must shift to the right to form more trans. "-)
I......0.250........0.145
C.......-x...........+x
E......0.250-x.....0.145+x
Substitute the E line into Kp expression and solve for x, 0.250-x and 0.145+x.
How did I know to make the C line -x......+x an not +x......-x; i.e., how do I know it will shift from cis to trans and NOT trans to cis.
I calculated Qp = 0.145/0.250 = 0.58 and compared that value with Kp. It is smaller than Kp and that tells me the numerator is too small and the denominator is too large so I know the reaction must shift to the right to form more trans. "-)
Answered by
Anonymous
.0898 .305
Answered by
Bro
You must solve for Q first to see which way the reaction proceeds.
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