Question

Consider the decomposition of the compound C₅H₆O₃as follows:

${\mathit{C}}_{\mathbf{5}}{\mathit{H}}_{\mathbf{6}}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathit{C}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathit{C}\mathit{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

When a 5.63-g sample of pure C₅H₆O₃(g) was sealed inan otherwise empty 2.50-L flask and heated to 200.⁰C,the pressure in the flask gradually rose to 1.63 atm andremained at that value. Calculate K for this reaction.

Short answer

The value of K for the given reaction will be .$6.74\times {10}^{-6}mo{l}^3/L^3$

Step-by-step solution

Step 1:Determinenumber of moles of gas present at equilibrium

Initial number of moles of C₅H₆O_3:

$5.63g{C}_5{H}_6{O}_3\times \frac{1mol{C}_5{H}_6{O}_3}{114.10g}=0.0493mol{C}_5{H}_6{O}_{3}initially$

Total number of moles of gaspresent at equilibrium

$\begin{array}{c}{n}_{total}=\frac{P_{total}V}{RT}\\ =\frac{1.63atm\times 2.50L}{\frac{0.08206atm}{Kmol}\times 473K}\\ =0.105mol\end{array}$

$\begin{array}{l}{C}_5{H}_6{O}_3\left(g\right)\rightleftharpoons C_2{H}_6\left(g\right)+3CO\left(g\right)\\ Initial:0.0493mol00\\ Change:-x+x+3x\\ Equil:0.0493-xx3x\end{array}$

Determine the value of the reaction constant

$\begin{array}{c}K=\frac{\left[C_2{H}_6\right]{\left[CO\right]}^3}{\left[C_5{H}_6{O}_3\right]}\\ K=\frac{\left[\frac{0.0186mol{C}_2{H}_6}{2.50L}\right]{\left[\frac{3\left(0.0186\right)molCO}{2.50L}\right]}^3}{\left[\frac{(0.0493-0.0186)mol{C}_5{H}_6{O}_3}{2.50L}\right]}\\ K=6.74\times {10}^{-6}mo{l}^3{L}^3\end{array}$