Question

Chemical thermodynamics"

Calculate A,H, A,S, and A,G using the following reaction:

MgCO3(s) + 2 HCl (1)→ MgCl2(s) + H2O(1)

Calculate the Gibbs energy and determine whether the reaction can proceed at temperatures of 293 and 1293 Κ.

Variant 19

Answer 1

To calculate the standard Gibbs free energy (ΔGº) at 298 K for the reaction 2 H₂(g) + O₂(g) → 2 H₂O(g), the formula ΔGº = ΔHº - TΔSº is used.

Step-by-step explanation:

To calculate the standard Gibbs free energy (ΔGº) at 298 K for the given reaction: 2 H₂(g) + O₂(g) → 2 H₂O(g), we need to use the equation: ΔGº = ΔHº - TΔSº, where ΔHº is the change in enthalpy, ΔSº is the change in entropy, and T is the temperature in Kelvin. We can obtain the values of ΔHº and ΔSº from the relevant information provided.

Using the given information, the ΔHº for the reaction is 571.6 kJ/mol and the ΔSº is 162.6 J/mol·K. Converting the temperature to Kelvin (298 K), we can now calculate the ΔGº at 298 K:

ΔGº = 571.6 kJ/mol - 298 K * (162.6 J/mol·K / 1000 J/kJ)

ΔGº = 571.6 kJ/mol - 48.0516 kJ/mol

ΔGº = 523.5484 kJ/mol

To determine if the reaction is spontaneous, we look at the sign of ΔGº. If ΔGº is negative, the reaction is spontaneous. In this case, ΔGº is 523.5484 kJ/mol, which is positive, indicating that the reaction is not spontaneous.

To calculate the ΔG at 330 K, we use the same formula: ΔG = ΔH - TΔS. However, we substitute the new temperature (330 K) and the same values of ΔHº and ΔSº:

ΔG = 571.6 kJ/mol - 330 K * (162.6 J/mol·K / 1000 J/kJ)

ΔG = 571.6 kJ/mol - 54.012 kJ/mol

ΔG = 517.588 kJ/mol

Comparing the ΔGº at 298 K (523.5484 kJ/mol) to the ΔG at 330 K (517.588 kJ/mol), we can see that the reaction becomes more spontaneous at higher temperatures as the ΔG value decreases. However, at both temperatures (293 K and 1293 K), the reaction is non-spontaneous because the ΔG values are positive.