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: What is K for a reaction if ∆G° =-269.9 kJ/mol at 25°C? (R = 8.314 J/mol ･ K)
A: Answer :- K = 2.0 × 1047 the equilibrium constant, K for the reaction = 2.0 × 1047…
Q: Determine ∆G° for a reaction when ∆G = -126.1 kJ/mol and Q = 0.043 at 298 K. (R = 8.314 J/mol ･ K)
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Q: At 460 K, ∆G = -15.00 kJ/mol for the reaction 2 A (g) → B (g). If the partial pressures of A and B…
A: Answer: ∆G° for this reaction is 3.42 kJ/mol.
Q: For the gas-phase equilibrium A(g) + 2 B(g) ⇌ C(g) the initial partial pressures of A, B, and C are…
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Q: What is ∆G0 for the reaction CN- + H2O = HCN + OH-? Ka for HCN = 4.0x10-10. R = 8.314 J/(mol*K)
A: "Gibbs free energy" is an estimation whether a system is able to do work or not. It is the energy…
Q: Determine ∆G° for a reaction when ∆G = -88.9 kJ/mol and Q = 0.043 at 298 K. (R = 8.314 J/mol ･ K)
A: The objective of this question is to calculate the standard Gibbs free energy change (∆G°) for a…
Q: a) image attached b) Calculate ΔG∘ at 296 ∘C c) tell whether the reaction has an equilibrium…
A:
Q: Determine ∆G° for a reaction when ∆G = -122.6 kJ/mol and Q = 0.043 at 298 K. (R = 8.314 J/mol ･ K)
A: The free energy change of the reaction is = -122.6 kJ/mol The temperature of the reaction is = 298 K…
Q: Determine the equilibrium constant for the following reaction at 298 K to one decimal point. SO3(g)…
A: We know that change in the standard Gibbs free energy is equal to the product of gas constant ,…
Q: What is ΔG° at 25 ºC for a process that has ΔH° = −35.4 kJ/mol and ΔS° = −153 J/mol·K? 45.6 kJ/mol…
A: Given: ΔH° = −35.4 kJ/mol ΔS° = −153 J/mol·K Temperature T = 25 ºC = 273 + 25 = 298 K
Q: What is ∆G for a reaction where ∆G° = -4.5 kJ/mol and Q = 3.0 at 295 K? (R = 8.314 J/mol ･ K)
A: The spontaneity and feasibility of reaction can be predicted by the Gibb’s free energy of that…
Q: Determine ∆G° for the reaction 2 NO₂(g) ⇌ N₂O₄(g) if K= 6.94 at 25.0 °C. (R = 8.314 J/mol･K)
A: Relation of gibbs free energy with equilibrium constant is given as: ∆G = ∆Go + RTlnK At…
Q: a) What is K for a reaction if ∆G° =151.6 kJ/mol at 25.0 °C? (R = 8.314 J/mol ･ K) b) For the…
A: We know- ΔG = ΔG° + RT ln QandΔGo=-RTlnK now we have to rearrange the equation- ΔG = ΔG° + RT ln QΔG…
Q: Determine ∆G° for a reaction when ∆G = -163.8 kJ/mol and Q = 0.043 at 298 K. (R = 8.314 J/mol ･ K)
A: The standard Gibb's free energy change and reaction quotient have the following relationship. ΔG =…
Q: Determine ∆G° for a reaction when ∆G = -181.2 kJ/mol and Q = 0.043 at 298 K. (R = 8.314 J/mol ･ K)
A: Please find the attachment
Q: What is K for a reaction if ∆G° =-164.6 kJ/mol at 25 °C (or 298 K)? (R = 8.314 J/mol ･ K)
A: Thermodynamics can be defined as a branch of chemistry in which we deal with the study of work, heat…
Q: What is K for a reaction if ∆G° =-369.4 kJ/mol at 25.0 °C? (R = 8.314 J/mol ･ K)
A: 1- First converted ∆G° unit kJmol-1 to Jmol-1 ∆G° = -369. 4 kJmol-1 We know that, 1kJ = 1000 J…
Q: What is K for a reaction if ∆G° =-232.9 kJ/mol at 25°C (or 298 K)? (R = 8.314 J/mol ･ K)
A: We know that ∆G° = - RT lnK
Q: What is K for a reaction if ∆G° =-292.8 kJ/mol at 25°C? (R = 8.314 J/mol ･ K)
A: Interpretation: The value of K for a reaction, if ∆G° = -292.8 kJ/mol at 25°C is to be determined.
Q: What is K for a reaction if ∆G° =-51.2 kJ/mol at 25.00 °C? (R = 8.314 J/mol ･ K)
A: The objective of this question is to calculate the equilibrium constant (K) for a reaction given the…
Q: Determine ∆G° for a reaction when ∆G = -195.4 kJ/mol and Q = 0.064 at 325 K. (R = 8.314 J/mol ･ K)
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: What is K for a reaction if ∆G° =-195.9 kJ/mol at 25°C? (R = 8.314 J/mol ･ K)
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Q: Given the reaction AB and using ∆Go = -RTln(Keq). (Useful calculations: ln1=0, ln1=positive; R =…
A: Answer: Ratio of molar concentrations of products and reactants is equal to reaction quotient and…
Q: The potassium-ion concentration in blood plasma is about 5.0 * 10-3 M, whereas the concentration in…
A:
Q: What is K for a reaction if ∆G° =-273.8 kJ/mol at 25°C? (R = 8.314 J/mol ･ K)
A: Standard Gibbs free energy is related with Equilibrium constant,K as - ∆G° = - RT ln K Where - ∆G°…
Q: What is K for a reaction if ∆G° =-254.1 kJ/mol at 25°C? (R = 8.314 J/mol ･ K)
A: The most usual method to state about reaction's spontaneity is comparing its Gibbs energy or its…
Q: What is ∆G for a reaction where ∆G° = -4.5 kJ/mol and Q = 2.5 at 295 K? (R = 8.314 J/mol ･ K)
A: According to thermodynamics equation, dG = dG° + RTlnQ On solving above equation we get,
Q: Determine ΔG0 at 298 K for N2(g) + 3 H2(g) <---> 2NH3(g) given Substance ΔH0f (kJ/mol)…
A: The given reaction is, N2(g) + 3H2(g) → 2NH3(g) The enthalpy change of the reaction can be…
Q: What is ∆G° for the reaction 3 O₂(g) → 2 O₃(g) at 25°C if K = 6.1 x 10⁻⁵⁸? (R = 8.314. J/mol ･ K)
A: Relation between ∆G0 and Equilibrium Constant a K is ∆G0=-RTlnK where ∆G0 is Gibbs free energy Ris…
Q: What is K for a reaction if ∆G° =173.2 kJ/mol at 25.0 °C? (R = 8.314 J/mol ･ K)
A: Given, Standard gibbs free energy (∆G°) = 173.2 kJ/mol = 173.2 × 1000 J/mol = 173200 J/mol…
Q: What is K for a reaction if ∆G° =-180.3 kJ/mol at 25°C (or 298 K)? (R = 8.314 J/mol ･ K)
A:
Q: What is K for a reaction if ∆G° =-64.2 kJ/mol at 25.0 °C? (R = 8.314 J/mol ･ K)
A:
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: What are the two organic compounds required to form the ester shown here? A) ethanoic acid and…
A: Here we are required to find the Starting material required to prepare ester as well as name the…
Q: Determine ∆G° for a reaction when ∆G = -152.3 kJ/mol and Q = 0.043 at 298 K. (R = 8.314 J/mol ･ K)…
A: We have to find value of ∆G°
Q: What is K for a reaction if ∆G° =-171.1 kJ/mol at 25°C? (R = 8.314 J/mol ･ K)
A: Gibbs' free energy and equilibrium constant has the following relation ∆G0 =-RT lnKeqm Or, Keqm =…
Q: What is ∆G for a reaction where ∆G° = -4.5 kJ/mol and Q = 1.6 at 295 K? (R = 8.314 J/mol ･ K)
A: Given: Standard Gibbs free energy change, ∆G° = -4.5 kJ/mol Reaction quotient, Q = 1.6 Temperature,…
Q: What is ∆G for a reaction where ∆G° = -4.5 kJ/mol and Q = 2.0 at 295 K? (R = 8.314 J/mol ･ K)
A:
Q: The equilibrium constant for a chemical reaction is Kc = 243 at 25°C. What is the standard free…
A: Given : Equilibrium constant, K = 243 And temperature = 25 oC
Q: A 0.15 M aqueous solution of a weak acid (HA) has a pH of 4.55 at 25 C. What is the ∆G for the…
A:
Q: What is K for a reaction if ∆G° =-364.8 kJ/mol at 25.0 °C? (R = 8.314 J/mol ･ K)
A: Given, Standard gibbs free energy (∆G°) = -364.8 kJ/mol = -364.8 × 1000 J/mol = -364800 J/mol…
Q: The ∆H° for a reaction is -46.8 kJ mol⁻¹ and ∆S° is -0.152 kJ mol⁻¹ K⁻¹. Calculate the ∆G° for this…
A: ∆H° = -46.8 kJ mol⁻¹∆S° = -0.152 kJ mol⁻¹ K⁻¹T = 298 K∆G° = ?Note: Relationship between enthalpy…
What is K for a reaction if ∆G° =-216.0 kJ/mol at 25°C (or 298 K)? (R = 8.314 J/mol ･ K)
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