Question

Can you help with part B

Part A

Acetylene, C2H2, can be converted to ethane, C2H6, by a process known as hydrogenation. The reaction is

C2H2(g)+2H2(g)⇌C2H6(g)

Given the following data at standard conditions (all pressures equal to 1 atm and the common reference temperature 298 K), what is the value of Kp for this reaction?

Substance ΔG∘f
(kJ/mol)
C2H2(g) 209.2
H2(g) 0
C2H6(g) −32.89
Express your answer using two significant figures.

Kp =
2.7×1042

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Correct

Based on the magnitude of K, we know that this reaction has practically gone to completion at equilibrium.

Standard versus Nonstandard Conditions

In Part A, we saw that ΔG∘=−242.1 kJ for the hydrogenation of acetylene under standard conditions (all pressures equal to 1 atm and the common reference temperature 298 K). In Part B, you will determine the ΔG for the reaction under a given set of nonstandard conditions.

Answer 1

`\delta G= -261.2kg` for the reaction under a given set of nonstandard conditions.

Step-by-step explanation:

`C_(2)H_6(g) + 2H_2(g)\rightleftharpoons C_2H_6(g)`

`Q_p = (P_c_2H_6)/(P_c_2H_2*P_H_2)`

=
`(3.25*10^-2)/(4.25*4.15)`

`Q_p` =
`4.44 * 10^-4`

`\delta G =\delta G^0 + RTlnQ_P`

=
`-242.1+8.314*10^-3*298*ln(4.44*10^-4)`

=
`\delta G= -261.2kg`

So,
`\delta G= -261.2kg`