DeltaG= H2=0 C2h4=68.2 C2h6=-32.8 eVG chegg Study iGuided ( xysowiv2IOnine teach

Question

DeltaG= H2=0 C2h4=68.2 C2h6=-32.8 eVG chegg Study iGuided ( xysowiv2IOnine teachr.VD Thermodyname Data × do?locator- assignment-take&takeAssignmentSessionlocator; assignment take keAssignm Use the Refereaces to access important values if needed for this question Consider the reaction Using the standard thermodynamic data in the tables linked above, calculate A Gr for this reaction at 298 15K if the pressure of each gas is 35.18 mm Hg ANSWERmol Retry Entre Group 8 more group attempts remaining Submit Answer

Explanation / Answer

Given:

Gof(H2(g)) = 0.0 KJ/mol

Gof(C2H4(g)) = 68.2 KJ/mol

Gof(C2H6(g)) = -32.8 KJ/mol

Balanced chemical equation is:

H2(g) + C2H4(g) ---> C2H6(g)

delta Go rxn = 1*Gof(C2H6(g)) - 1*Gof( H2(g)) - 1*Gof(C2H4(g))

delta Go rxn = 1*(-32.8) - 1*(0.0) - 1*(68.2)

delta Go rxn = -101 KJ

delta Go rxn = -101000 J

Pressure of each gas is,

p = 35.18 mmHg

= 35.18 / 760 atm

= 0.04629 atm

Now use:

delta G rxn = delta Go rxn + R*T*ln Q

delta G rxn = delta Go rxn + R*T*ln (p(C2H6) / p(H2)p(C2H4))

delta G rxn = delta Go rxn + R*T*ln (p/p*p)

delta G rxn = delta Go rxn + R*T*ln (1/p)

delta G rxn = -101000 + 0.0821*298.15*ln (1/0.04629)

delta G rxn = -101000 + 0.0821*298.15*3.0728

delta G rxn = -101000 + 75.22

delta G rxn = -100.924 J

delta G rxn = -100.9 KJ

Answer: -100.9 KJ

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