Interpretation:
The spontaneous and non-spontaneous processes for the reactions are to be determined.
Concept introduction:
Gibbs free energy is defined as the energy that can be used to get work done. The abbreviation for Gibbs Free Energy is
Gibbs free energy is defined as the energy that can be used to get work done. It is denoted a
The standard free energy, abbreviated as
The Gibbs free energy change is calculated by using the expression as:
Here,
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Chemistry
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- At 276.7 oC, the Kp for the reaction 2 A (g) + 1 B (g) ↔ 5 AB is Kp = 3.07⋅10-2 What is Keq?arrow_forwardConsider the dissociation of methane, CH4(g), into the elements H2(g) and C(s, graphite). (a) Given that ΔfH⦵(CH4,g) = −74.85 kJ mol−1 and that ΔfS⦵ = −80.67 J K−1 mol−1 at 298 K, calculate the value of the equilibrium constant at 298 K. (b) Assuming that ΔrH⦵ is independent of temperature, calculate K at 50 °C. (c) Calculate the degree of dissociation, α, of methane at 298 K and a total pressure of 0.010 bar. (d) Without doing any numerical calculations, explain how the degree of dissociation for this reaction will change as the pressure and temperature are varied.arrow_forwardSelect the reaction for which K, = Kc. 2 CO, (g) + 2 CF,(g) =4 COF, (g) NH,NO, (s) = N,O(g) + 2H,O(g) 2 H, S(g) + SO, (g) = 3 S(s) + 2 H,O(g) O 2 Na, O, (s) + 2CO,(g) =2 Na,CO;(s) + 0,(g)arrow_forward
- (a) Calculate K at 298 K for the following reaction: 1 NO(g) + O₂(g) ⇒ NO₂(g) K = 1.741 NO(g) 86.60 NO₂(g) 51 O₂(g) 0 Part 2 out of 2 AG () f mol AG⁰ = x 10 (b) Use the equilibrium constant to calculate AGº (in kJ) at 298 K for the following reaction: 2HCl(g) → H₂(g) + Cl₂(g) K = 3.89 × 10-34 at 298 K x 10 kJarrow_forwardCalculate the equilibrium constant for the following reaction at 25o C, given that change in standard Gibbs free energy (f) of O3 (g) is 163.4 kJ/mol. 2O3(g) -> 3O2(g)arrow_forwardDetermine the equilibrium constant K_{p} at 25°C for the reaction N 2 (g) 3H 2 (g) cong2NH 3 (g); [ Delta G^ f(NH 3 (g))=-16.6 kJ/mol]arrow_forward
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