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3 years ago

First answer will be brainliest!!

Chemistry

1 answer:

igomit [66]3 years ago

8 0

I think it is rarefaction. But im not sure

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3 years ago

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3 years ago

Consider the equilibrium

vladimir1956 [14]

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.

3 0

4 years ago

Please help me !!!!!

Margarita [4]

I cant read it, please make it focused

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3 years ago

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