# For the reaction

2CO2(g) + 5H2(g)C2H2(g) + 4H2O(g)

H° = 46.5 kJ and S° = -124.8 J/K

The standard free energy change for the reaction of 1.75 moles of CO2(g) at 328 K, 1 atm would be ___ kJ

This reaction is (reactant, product) ____ favored under standard conditions at 328 K.

Assume that delta H and delta S are independent of temperature.

## I'm confused by the problem. Is that So and Ho for the reaction. If so, then

dG = dH - TdS.

Products are favored if dG is -.

## To find the standard free energy change (ΔG°) for the reaction, we can use the formula:

ΔG° = ΔH° - TΔS°

where ΔH° is the standard enthalpy change and ΔS° is the standard entropy change. T represents the temperature in Kelvin.

Given:

ΔH° = 46.5 kJ

ΔS° = -124.8 J/K

T = 328 K

First, we need to convert ΔS° to kJ/K. Since 1 kJ = 1000 J, the conversion factor will be 1/1000. Therefore:

ΔS° = -124.8 J/K * (1 kJ/1000 J) = -0.1248 kJ/K

Now, we can substitute the values into the equation:

ΔG° = 46.5 kJ - (328 K)(-0.1248 kJ/K)

= 46.5 kJ + 40.9152 kJ

= 87.4152 kJ

Therefore, the standard free energy change for the reaction of 1.75 moles of CO2(g) at 328 K and 1 atm is approximately 87.4152 kJ.

To determine whether the reaction is reactant or product favored under standard conditions at 328 K, we need to check the sign of ΔG°:

If ΔG° < 0, the reaction is product-favored.

If ΔG° > 0, the reaction is reactant-favored.

In this case, since ΔG° is positive (87.4152 kJ), the reaction is reactant-favored under standard conditions at 328 K.

## To calculate the standard free energy change (ΔG°) for a reaction, we can use the equation:

ΔG° = ΔH° - T ΔS°

where ΔH° is the standard enthalpy change, ΔS° is the standard entropy change, and T is the temperature in Kelvin.

Given:

ΔH° = 46.5 kJ

ΔS° = -124.8 J/K

Temperature (T) = 328 K

First, let's convert the given entropy change to kilojoules:

ΔS° = -124.8 J/K = -0.1248 kJ/K

Now, substitute these values into the equation to calculate ΔG°:

ΔG° = ΔH° - T ΔS°

ΔG° = 46.5 kJ - (328 K x -0.1248 kJ/K)

ΔG° = 46.5 kJ + 40.8512 kJ

ΔG° = 87.3512 kJ

Therefore, the standard free energy change for the reaction of 1.75 moles of CO2(g) at 328 K, 1 atm is 87.3512 kJ.

To determine whether the reaction is favored or not under standard conditions at 328 K, we compare the sign of ΔG° to zero. If ΔG° < 0, the reaction is favored (spontaneous) under standard conditions. If ΔG° > 0, the reaction is not favored (non-spontaneous) under standard conditions.

Since the value of ΔG° is positive (87.3512 kJ), the reaction is NOT favored under standard conditions at 328 K.