Answer: the standard change in Gibbs free energy of the reaction at 25 ∘C is 6.24 kJ/mol.
Explanation: One can employ the equation in order to calculate the standard Gibbs free energy change (ΔG°) for the reaction at a temperature of 25 degrees Celsius.
The standard free energy change, ΔG°, can be expressed as the negative product of the universal gas constant (R), temperature (T) and the natural logarithm of the equilibrium constant (K).
The equilibrium constant is denoted by K and the gas constant R has a value of 8.314 J/mol·K, while T stands for temperature in Kelvin, which is equivalent to 298 K for 25 ∘C.
K can be determined by utilizing the partial pressures at equilibrium.
K equals the square of the product of PC and the cube of PD divided by the cube of PA multiplied by the fourth power of PB.
After replacing the provided values, the result obtained is:
The value of K is obtained by raising 4.36 atm to the power of 2, and 4.70 atm to the power of 3, and then dividing that by the product of 4.62 atm to the power of 3 and 4.36 atm to the power of 4.
The numeric value of K is 0.0786.
We can now compute the value of ΔG° by using the available data.
The change in Gibbs energy under standard conditions is equal to the negative product of gas constant, temperature and natural logarithm of equilibrium constant.
The standard Gibbs free energy change is determined by multiplying the constant of gas by the temperature and natural logarithm of the equilibrium constant.
The value of ΔG° can be expressed as - (8.314 J/mol·K) multiplied by 298 K and -2.547.
The standard free energy change is either 6,237 joules per mole or 6.24 kilojoules per mole.