Answer:
[tex]K_{eq}=1.02x10^{-4}[/tex]
Explanation:
Hello,
In this case, the first step is to compute the initial moles of C₅H₆O₃ as shown below:
[tex]5.63gC_5H_5O_3*\frac{1molC_5H_5O_3}{114gC_5H_5O_3}=0.0494molC_5H_5O_3[/tex]
After that, by knowing that the final pressure is 1.63 atm, one computes the total moles at the equilibrium as follows:
[tex]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[/tex]
Then, by knowing the moles at the equilibrium considering the change "[tex]x[/tex]", which yields to:
[tex]\ \ \ \ \ 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[/tex]
The total moles at the equilibrium turn out:
[tex]n_{total}^{eq}=0.0494mol-x+x+3x[/tex]
By solving for "[tex]x[/tex]", we've got:
[tex]3x=0.105mol-0.0494mol\\x=\frac{0.0556mol}{3}\\x=0.0185mol[/tex]
Finally, the equilibrium constant is:
[tex]K_{eq}=\frac{(x)(3x)^3}{0.0494-x}=\frac{(0.0185mol)(3*0.0185mol)^3}{0.0494mol-0.0185mol}=1.02x10^{-4}[/tex]
Best regards.
