The equation is: CH4(g) + CO2(g) <--> 2CO(g) + 2H2(g)
(the <--> stands for at equilibrium)
A 1.00-L flask is filled with 0.30 mol of CH4 and 0.40 mol of CO2, and allowed to come to equilibrium. At equilibrium, there are 0.20 mol of CO in the flask. What is the value of Kc, the equilibrium constant, for the reaction?
A. 1.2
B. 0.027
C. 0.30
D. 0.060
E. 3.0
2007-04-23 09:18:25 · 3 answers · asked by delilah 2 in Science & Mathematics ➔ Chemistry
Could you show me how you did it, too?
2007-04-23 09:27:46 · update #1
The equilibrium constant is given by
[CO]2 [H2]2 / [CH4][CO2] = Kc
From the reaction, to form 0.20 moles of CO, 0.10 moles each of CH4 and CO2 have been reacted. Also, if 0.20 moles of CO are formed, so are 0.20 moles of H2 gas.
THEN: [0.2]2 [0.2]2 / [0.2][0.3] = Kc
Kc = 0.0016/0.06 = 0.0.027
This corresponds most closely to B
2007-04-23 09:36:50 · answer #1 · answered by cattbarf 7 · 2⤊ 0⤋
D
2007-04-23 16:25:34 · answer #2 · answered by BillyBoy 2 · 0⤊ 1⤋
why can't you people do ur own homework?? GEEZ
2007-04-23 16:27:25 · answer #3 · answered by determined_ladii 4 · 0⤊ 3⤋