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Marrrta [24]
3 years ago
8

Consider the equilibrium

Chemistry
1 answer:
vladimir1956 [14]3 years ago
3 0

Answer:

Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}

\Delta _rG=1.01x10^5J/mol

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar

Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}

Finally, the Gibbs free energy for the reaction at 298.15K is:

\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol

Best regards.

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