Q1) For the reaction, A �¨ B,
ƒ¢H�‹ = +11.00 kJ mol-1 and Keq is 12.50
Calculate i) ƒ¢G�‹ and ii) ƒ¢S�‹ at 25 �‹C.
[0 �‹C = 273.15 K; R = 8.314 J K-1 mol-1]
To calculate ƒ¢G�‹ (change in Gibbs free energy) and ƒ¢S�‹ (change in entropy) for the given reaction at 25 �‹C, we can use the following equations:
i) ƒ¢G�‹ = -RTln(Keq)
ii) ƒ¢G�‹ = ƒ¢H�‹ - Tƒ¢S�‹
Let's solve these equations step by step.
i) Calculate ƒ¢G�‹:
In the equation ƒ¢G�‹ = -RTln(Keq), we have the values of R and T. R is the universal gas constant (8.314 J K-1 mol-1), and T is the temperature in Kelvin (25 �‹C = 25 + 273.15 K).
Now, substitute the values into the equation:
ƒ¢G�‹ = - (8.314 J K-1 mol-1)(25 + 273.15 K)ln(12.50)
Use the natural logarithm (ln) function to calculate the logarithm value.
ii) Calculate ƒ¢S�‹:
In the equation ƒ¢G�‹ = ƒ¢H�‹ - Tƒ¢S�‹, we already know the values of ƒ¢G�‹ and T (25 �‹C = 25 + 273.15 K). We can rearrange the equation to solve for ƒ¢S�‹:
ƒ¢S�‹ = (ƒ¢H�‹ - ƒ¢G�‹) / T
Substitute the known values:
ƒ¢S�‹ = (11.00 kJ/mol - (-RTln(Keq))) / (25 + 273.15 K)
Convert the temperature into Kelvin.
Now, solve the equation using the calculated ƒ¢G�‹ value.
By following these steps, you can calculate ƒ¢G�‹ and ƒ¢S�‹ for the given reaction at 25 �‹C.