Decomposition of C5H6O3 equilibrium

[i_love_science]

Homework Statement

Consider the decomposition of the compound C5H6O3 as follows:
C5H6O3(g) --> C2H6(g) + 3CO(g)
When a 5.63-g sample of pure C5H6O3(g) was sealed in an otherwise empty 2.50-L flask and heated to 200°C, the pressure in the flask gradually rose to 1.63 atm and remained at that value. Calculate Kc for this reaction.

Relevant Equations

equilibrium expression

My solution:

partial pressure of C5H6O3 = mRT/MV = (5.63 g)(0.08206 L*atm/mol*K)(473 K) / (114.098 g/mol)(2.50 L) = 0.766 atm

equilibrium partial pressure of C5H6O3 = 0.766 - x
equilibrium partial pressure of C2H6 = x
equilibrium partial pressure of CO = 3x
total pressure = 0.766 atm - x + x + 3x = 1.63 atm
x = 0.288 atm
Kp = (0.288 atm)(3*0.288 atm)^3 / (0.766-0.288) = 0.389
Kc = Kp / (RT)^(delta n) = 0.389 / [(0.08206 L*atm/mol*K)(473 K)]^1 = 0.01

My answer was wrong. The correct answer is 6.74*10^-6, and was found by doing all the calculations directly in terms of moles and concentration. Could anyone explain where I went wrong? Thanks.

 

[ergospherical]

Nearly all fine, except for the final conversion from $K_p$ to $K_c$. Notice that\begin{align*}

K_c = \dfrac{[\mathrm{CO}]^3 [\mathrm{C_2H_6}]}{[\mathrm{C_5 H_6 O_3}]} &= \dfrac{\left(\dfrac{n_{\mathrm{CO}}}{V}\right)^3 \left(\dfrac{n_{\mathrm{C_2H_6}}}{V}\right)}{\left(\dfrac{n_{\mathrm{C_5 H_6 O_3}}}{V}\right)} \times \dfrac{(RT)^4}{(RT)^4} \\ \\

&= \dfrac{\left(\dfrac{n_{\mathrm{CO}} RT }{V}\right)^3 \left(\dfrac{n_{\mathrm{C_2H_6}} RT}{V}\right)}{\left(\dfrac{n_{\mathrm{C_5 H_6 O_3}} RT}{V}\right)} \times \dfrac{1}{(RT)^3} = \dfrac{K_p}{(RT)^3}

\end{align*}In other words, in the formula $K_c = K_p / (RT)^{\Delta n}$ you should have $\Delta n = 3$, which you can tell directly from the reaction equation.