Question

Consider the decomposition of the compound C5H6O3 as follows: C5H6O3(g)  C2H6(g) + 3CO(g) A 5.63 g sample of pure C5H6O3(g) was placed in an evacuated 2.50 L flask and heated to 200.ºC. At equilibrium, the pressure in the flask was 1.63 atm. Calculate K for this reaction.

Answer 1

Answer:

$K_{eq}=1.02x10^{-4}$

Explanation:

Hello,

In this case, the first step is to compute the initial moles of C₅H₆O₃ as shown below:

$5.63gC_5H_5O_3*\frac{1molC_5H_5O_3}{114gC_5H_5O_3}=0.0494molC_5H_5O_3$

After that, by knowing that the final pressure is 1.63 atm, one computes the total moles at the equilibrium as follows:

$n_{total}^{eq}=\frac{P_{total}^{eq}V}{RT}=\frac{1.63atm*2.50L}{0.082\frac{atm*L}{mol*K}*473.15K} =0.105mol$

Then, by knowing the moles at the equilibrium considering the change "$x$", which yields to:

$\ \ \ \ \ C_5H_6O_3(g) \leftrightarrow C_2H_6(g) + 3CO(g)\\I\ \ \ \ \ 0.0494mol\ \ \ \ \ \ 0mol \ \ \ \ \ \ \ \ 0mol\\C\ \ \ \ \ \ -x\ \ \ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3x\\E\ \ 0.0494mol-x\ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ 3x$

The total moles at the equilibrium turn out:

$n_{total}^{eq}=0.0494mol-x+x+3x$

By solving for "$x$", we've got:

$3x=0.105mol-0.0494mol\\x=\frac{0.0556mol}{3}\\x=0.0185mol$

Finally, the equilibrium constant is:

$K_{eq}=\frac{(x)(3x)^3}{0.0494-x}=\frac{(0.0185mol)(3*0.0185mol)^3}{0.0494mol-0.0185mol}=1.02x10^{-4}$

Best regards.