10. What is ∆G for the decomposition of CaCO3 at 298K and a partial pressure of CO2 of 4*10-4 bar? ∆Gf0 for CaCO3(s), CaO(s) and CO2(g) are -1129 kJ/mol, -604 kJ/mol and -394 kJ/mol, respectively. CaCO3(s) → CaO(s) + CO2(g) A. 0 B. -131 kJ/mol C. 112 kJ/mol D. 131 kJ/mol E. 150 kJ/mol   (Correct answer is C, I'm just looking for an explanation!).

Q: For the reaction A (g) → 3 B (g), Kp = 0.411 at 298 K. What is the value of ∆G for this reaction at…

A: Ans is -12.41kj/mol

Q: The equilibrium constant KP is 6.0 x 1015 at 298 K for the reaction:  2H2S (g) + SO2 (g)  ⇆  3S (s)…

A: Given reaction is : 2H2S (g) + SO2 (g)  ⇆  3S (s) + 2H2O (g)  The equilibrium constant KP = 6.0 x…

Q: 415 K, ∆G = -21.9 kJ/mol for the reaction 2 A (g) → B (g). If the partial pressures of A and B are…

A: ∆G = ∆G° + RTlnQ  T = temperature Q = reaction quotient

Q: Determine ∆G° for the reaction 3 N₂(g) + 2 O₃(g) → 6 NO(g) at 223°C.   ∆H°= +257.7 kj/mol ∆S°=…

A: Gibbs Free energy (G) : The energy associated with a chemical reaction that can be used to do work.…

Q: Calculate ∆S° for the following reactions in J K-1 a. 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) b.…

A: The standard molar entropy of the compounds in reaction (a) can be listed as follows: ∆Soethane =…

Q: At 1120 K, ∆G° = 47.1 kJ/mol for the reaction 3 A (g) + B (g) →2 C (g). If the partial pressures of…

A:

Q: Use data in the table below to establish at 298 K for the reaction: 2 NaHCO3 (s) → Na2CO3 (s) +…

A: Here we are required to find the equilibrium constant .

Q: Define spontaneity of a reaction in terms of ∆G. Determine if a process is spontaneous based on…

A:

Q: At 560 K, ∆G° = -28.9 kJ/mol for the reaction A (g) + 2 B (g) → 3 C (g). If the partial pressures of…

A:

Q: At 1120 K, ∆G° = 67.3 kJ/mol for the reaction 3 A (g) + B (g) →2 C (g). If the partial pressures of…

A:

Q: The neutralization of a strong acid and strong base has an enthalpy change, ∆H°, = -55.9 kJ/mol. The…

A: The relationship between ∆S°, ∆H° and ∆G° is given by  ∆G° = ∆H° - T∆S° where T = temperature in…

Q: The neutralization of a strong acid and strong base has an enthalpy change, ∆H°, = -55.9 kJ/mol. The…

A: The relationship between the ΔG0 and equilibrium constant is given by  RTln(K) = -ΔG0  where R = gas…

Q: For the reaction 2 A (g) → B (g), Kp = 0.00271 at 298 K. When ∆G = 8.35 kJ/mol, what is the partial…

A:

Q: When the reaction FeO (s) + CO (g) ↔ Fe (s) + CO2 (g) takes place at 600 K, , it is observed that a)…

A: a).  = equilibrium constant at 600°C =  0.900

Q: Consider the following reaction at 25 °C: 3 NiO(s) + 2 NH₃(g) → 3 Ni(s) + N₂(g) + 3 H₂O(g) If ∆G°…

A: The amount of energy related to a chemical reaction which can be used to extract some useful work is…

Q: At 560 K, ∆G° = -24.7 kJ/mol for the reaction A (g) + 2 B (g) → 3 C (g). If the partial pressures of…

A:

Q: For the following reaction 2NO₂(g) ⇌ N₂O₄(g) ∆H° = -58.02 kJ/mol and ∆S° = -176.6 J/mol･K. What is…

A: The reaction is –                    2NO2g ⇔N2O4g Pressure        0.2                0.2Qp…

Q: At 1275 K, ∆G° = 41.1 kJ/mol for the reaction 3 A (g) + B (g) → 2 C (g). If the partial pressures of…

A:

Q: Keq

A:

Q: Calculate the ∆G in kJ/mol for a reaction at room temperature (~25.00°C) that has a Keq = 1000.0.…

A: We have to calculate the ∆G.

Q: At 560 K, ∆G° = -18.1 kJ/mol for the reaction A (g) + 2 B (g) → 3 C (g). If the partial pressures of…

A: Gibbs free energy of the chemical reaction can be expressed as, ∆G=∆Go+RTlnQ where Q is the reaction…

Q: At 560 K, ∆G° = -23.5 kJ/mol for the reaction A (g) + 2 B (g) → 3 C (g). If the partial pressures of…

A:

Q: At 1120 K, ∆G° = 78.3 kJ/mol for the reaction 3 A (g) + B (g) →2 C (g). If the partial pressures of…

A: Given,   3 A (g) + B (g) →2 C (g) at 1120 K, ∆G° = 78.3 kJ/mol and the partial pressures of A, B,…

Q: At 1125 K, ∆G° = 28.4 kJ/mol for the reaction 3 A (g) + B (g) → 2 C (g). If the partial pressures of…

A:

Q: At 560 K, ∆G° = -38.9 kJ/mol for the reaction A (g) + 2 B (g) → 3 C (g). If the partial pressures of…

A: The question is based on the concept of equilibrium. we have to calculate Gibbs free energy change…

Q: Consider the following reaction at 25 °C: Ba(OH)₂･8 H₂O(s) + 2 NH₄Cl(s) → BaCl₂(s) + 2 NH₃(g) + 10…

A: Ba(OH)₂.8 H₂O(s) + 2 NH₄Cl(s) → BaCl₂(s) + 2 NH₃(g) + 10 H₂O(l) ∆G° = -14.1 kJ/mol The pressure of…

Q: At 1125 K, ∆G° = 43.2 kJ/mol for the reaction 3 A (g) + B (g) → 2 C (g). If the partial pressures of…

A: We have to calculate free energy for the reaction  3 A (g) + B (g) → 2 C (g)

Q: Under what temperature range would the following reaction be spontaneous?   N2(g) + 3 H2(g) ⇌ 2…

A: The given problem can be solved by using the formula given below as; ∆G°=∆H° - T∆S°…

Q: What is ∆G° for the reaction Br₂(g) → 2 Br(g) at 25.0 °C if K = 4.6 x 10⁻²⁹? (R = 8.314 J/mol･K) in…

A: Given reaction: Br₂(g) → 2 Br(g) K = 4.6 x 10⁻²⁹ We have to calculate ∆G° for the reaction.

Q: At 1120 K, ∆G° = 66.3 kJ/mol for the reaction 3 A (g) + B (g) →2 C (g). If the partial pressures of…

A:

Q: At 560 K, ∆G° = -39.9 kJ/mol for the reaction A (g) + 2 B (g) → 3 C (g). If the partial pressures of…

A: Equilibrium constant is defined as ratio of partial pressure of reactant to the partial pressure of…

Q: Dissolution of ionic compounds generally involves an increase in entropy (+ΔS°), but for a few…

A: entropy is measurement of randomness.As the randomness increase entropy also increase. In…

Q: For the following reaction H₂(g) + I₂(g) ⇌ 2HI(g) ∆H° = -10.64 kJ/mol and ∆S° = 21.48 J/mol･K. What…

A:

Q: At 560 K, ∆G° = -36.3 kJ/mol for the reaction A (g) + 2 B (g) → 3 C (g). If the partial pressures of…

A: Information about the question

Q: For the reaction A (g) → 3 B (g), Kp = 0.767 at 298 K. What is the value of ∆G for this reaction at…

A:

Q: At 1120 K, ∆G° = 46.5 kJ/mol for the reaction 3 A (g) + B (g) →2 C (g). If the partial pressures of…

A:

Q: At 415 K, ∆G = -30.9 kJ/mol for the reaction 2 A (g) → B (g). If the partial pressures of A and B…

A: ∆G = ∆G° + RTlnQ Q = reaction quotient R = gas constantT = temperature

Q: Consider the reaction: ICl(g) + Cl₂(g) → ICl₃(s). The ∆G° of the reaction is -17.09 kJ/mol.…

A: Given, ICl(g) + Cl₂(g) → ICl₃(s) ∆G° = -17.09 kJ/mol.

Q: At 298 K, the formation of ammonia has a negative ∆S0sys. Calculate ∆Ssurr               N2(g) +…

A: The balanced thermodynamic reaction given is  N2 (g) + 3 H2 (g) → 2 NH3 (g) ∆S°sys = – 197 J/K