Calculate ΔG° for the reaction below under standard conditions at 25°C. 2S(s) + 3O2(g) + 2H2O(l) → 2H2SO4(l) ΔH° = -1056 kJ/molΔS° = -505 J/molHint: Pay attention to units of ΔH° and ΔS°
Question
Calculate ΔG° for the reaction below under standard conditions at 25°C. 2S(s) + 3O2(g) + 2H2O(l) → 2H2SO4(l) ΔH° = -1056 kJ/molΔS° = -505 J/molHint: Pay attention to units of ΔH° and ΔS°
Solution
To calculate ΔG° for the reaction, we can use the Gibbs free energy equation:
ΔG° = ΔH° - TΔS°
Given in the problem, we have ΔH° = -1056 kJ/mol and ΔS° = -505 J/mol. However, we need to pay attention to the units. In the Gibbs free energy equation, ΔH° is in kJ/mol and TΔS° needs to be in the same units. Therefore, we need to convert ΔS° from J/mol to kJ/mol.
1 kJ = 1000 J, so ΔS° = -505 J/mol ÷ 1000 = -0.505 kJ/mol.
The temperature T is given as 25°C. We need to convert this to Kelvin because the standard unit of temperature in these equations is Kelvin.
T(K) = 25°C + 273.15 = 298.15 K.
Now we can substitute these values into the Gibbs free energy equation:
ΔG° = ΔH° - TΔS° ΔG° = -1056 kJ/mol - (298.15 K * -0.505 kJ/mol*K) ΔG° = -1056 kJ/mol - (-150.5 kJ/mol) ΔG° = -1056 kJ/mol + 150.5 kJ/mol ΔG° = -905.5 kJ/mol
So, the standard Gibbs free energy change for the reaction is -905.5 kJ/mol.
Similar Questions
51. What is the ΔH° at 298 K of the given reaction?*- 3.8 kJ/mol3.8 J/mol0Cannot defined
What is the ΔG° at room temperature of the reaction 2CO+O2→2CO2, if ΔH°=-128.3 kJ and ΔS°= - 159.5 J/K ?*- 80. 769 kJ80. 769 kJ- 175.831 kJ- 124.313 kJ
The value of ΔH° for the reaction below is - 790 kJ.2S(s) + 3O2(g) → 2SO3(g)The enthalpy change accompanying the reaction of 0.95 g of S is ________kJ.Question 4Select one:a.23b.-23c.-12d.12e.-790
What is ΔrG° (in kJ mol-1) for a redox reaction that has E°cell = -2.69 V at 25°C ? Assume 4 electrons are transferred in the balanced redox reaction.If needed, F = 96485 C/mol.Answer:Question 4
We have used the equilibrium expression, relating equilibrium constant and concentrations to calculate equilibrium constants. We can also calculate equilibrium concentrations using the equilibrium expression. Carefully sort out the known quantities and the unknown to solve the following.Determine the equilibrium concentration of the reactant, N2O4(g), for the following reaction system at 25°C : N2O4 (g) ⇋ 2NO2 (g) At 25°C, Kc = 0.040At equilibrium at 25°C, [NO2] = 0.107 mol/L.(write the equilibrium expression first, rearrange carefully)The equilibrium concentration of N2O4(g) in mol/L is Question 2Select one:a.2.68b.0.286c.3.49d.0.374e.0.0286
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.