This is a question on Non-standard states.

Calculate ¥ÄG605 for the following reaction under the given conditions. Use data obtained from the table below and assume that ¥ÄHo and ¥ÄSo do not vary with temperature.
C2H4(g) + H2O(g) ¡æ C2H5OH(g)

The ¥ÄHo and ¥ÄSo values as well as the partial pressure values are given for each substance in gas form. I am not sure how the partial pressure is invovled in calculating G value.

Thanks

nevermind, i figured this one out!

Gibbs free energy (G) is a thermodynamic quantity used to measure the spontaneity of a reaction under given conditions. It can be calculated using the equation:

ΔG = ΔH - TΔS

where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, ΔS is the change in entropy, and T is the temperature in Kelvin.

To calculate ΔG for the given reaction (C2H4(g) + H2O(g) ↔ C2H5OH(g)), you will need the ΔH and ΔS values for each substance involved in the reaction, as well as the temperature at which the reaction is taking place.

The partial pressure of each gas, mentioned in the question, plays a role in calculating the ΔG value through the equation:

ΔG = ΔG° + RTln(Q)

where ΔG° is the standard Gibbs free energy change for the reaction at standard conditions, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Q is the reaction quotient. The reaction quotient (Q) is calculated using the partial pressures of the gases involved in the reaction.

Since the given question mentions that the ΔH° and ΔS° values do not vary with temperature, it implies that the given values are at standard conditions (usually 298 K). This means that ΔG° is equivalent to ΔG at the given temperature.

To calculate ΔG, you need to calculate the reaction quotient (Q) using the partial pressures of the gases involved in the reaction. Then, you can use the equation mentioned earlier to calculate ΔG.

If you have already figured out the solution to the question, that's great! If you have any further questions or need assistance with any other topic, feel free to ask.