For the reaction C4H10 C4H6 + 2H2 the equilibrium constant Kc is 1.00x10^-6 at 600ºC. If 1.00 mol of C4H6 and 1.00 mol of H2 were introduced into a 1.00L flask at 600ºC, what would be the equilibrium concentration of each gas?

Question

For the reaction C4H10 <-> C4H6 + 2H2 the equilibrium constant Kc is 1.00x10^-6 at 600ºC. If 1.00 mol of C4H6 and 1.00 mol of H2 were introduced into a 1.00L flask at 600ºC, what would be the equilibrium concentration of each gas?

anonymous · Answer

I do it with RICE usually...So:
R) C4H10 <-> C4H6 + 2H2
I)  -----           1M         1M
C)
E)
I don't know how to get the C and E without the Initial Molarity of C4H10

anonymous · Answer

1.00 x 10^-6 = (1+x)(1+2x)^2/-x
Can you help me from here? @hartnn

hartnn · Answer

you sure thats correct equation ?? because you will get a really weird answer for thsi

anonymous · Answer

Pretty sure...
R) C4H10 <-> C4H6 + 2H2
I)  -----           1M         1M
C) -x                 +x          +2x
E) -x                 1+x           1+2x
Kc = [C4H6][H2]^2/[C4H10]

@ash2326 wanna try n help me if u know if this is right? <3

hartnn · Answer

this can become easier, if we approximate things,
like $\large 10^{-6} =0$ approximately, compared to 1

hartnn · Answer

so we would only have 
$(1+x)(1+2x)^2 =0$
which is very easy to solve

hartnn · Answer

$1+x = 0,  \quad x =...\ 1+2x = 0, \quad x=....$