question_answer 1)

\[{{K}_{p}}=0.04\]atm at 899 K for the quilibrium shown below. What is the equilibrium concentration (pressure) of \[{{C}_{2}}{{H}_{6}}\]when it is placed in flask at 4 atm pressure and allowed to come to equilibrium? \[{{C}_{2}}{{H}_{6}}(g)\rightleftharpoons {{C}_{2}}{{H}_{4}}(g)+{{H}_{2}}(g)\]  

Answer:

Information shadow: \[{{K}_{p}}\] of the reaction is 0.04 atm and initial pressure of \[{{C}_{2}}{{H}_{6}}\]was 4 atm. The pressure of \[{{C}_{2}}{{H}_{6}}\] at equilibrium is required. Problem solving strategy: \[\underset{\begin{smallmatrix} Initially \\ At\,equilibrium \end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\underset{\begin{smallmatrix} 4atm \\ (4-x) \end{smallmatrix}}{\mathop{{{C}_{2}}{{H}_{6}}(g)}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{{{C}_{2}}{{H}_{4}}(g)}}\,+\underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{{{H}_{2}}(g)}}\,\] The expression of \[{{K}_{p}}\] for the reaction: \[{{K}_{p}}=\frac{[{{p}_{{{C}_{2}}{{H}_{4}}}}][{{p}_{{{H}_{2}}}}]}{[{{p}_{{{C}_{2}}{{H}_{6}}}}]}\] Putting the values in above equation we can calculate the pressure of \[{{C}_{2}}{{H}_{6}}\] at equilibrium. Working it out: \[0.04=\frac{x\times x}{(4-x)}\] It gives a quadratic equation: \[{{x}^{2}}+0.04x-0.16=0\] \[x=\frac{-0.04\pm \sqrt{0.0016+4\times 1\times 0.16}}{2\times 1}\] \[=\frac{-0.04\pm 0.80}{2}\] \[=\frac{-0.04+0.80}{2}\] (taking positive value) \[=0.38\] \[{{p}_{{{C}_{2}}{{H}_{6}}}}=(4-x)=4-0.38=3.62atm\]