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: What is the equilibrium constant, K, for a reaction if ∆G° = -51.2 kJ/mol at 25.0 °C? (R = 8.314…
A:
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)
A:
Q: For the gas-phase equilibrium A(g) + 2 B(g) ⇌ C(g) the initial partial pressures of A, B, and C are…
A:
Q: What is ∆G for a reaction where ∆G° = -4.5 kJ/mol and Q = 4.0 at 295 K? (R = 8.314 J/mol ･ K)
A: The given problem can be solved by using the formula given below as; ∆G = ∆G° + 2.303 RT log Q…
Q: Determine ∆G° for the reaction N₂O₄(g) ⇌ 2 NO₂(g) (K= 0.144 at 298 K).
A: The objective of the question is to calculate the standard Gibbs free energy change (∆G°) for the…
Q: Calculate T (in K) given ΔG0= -89.9 kJ/mol and ΔH0= 83.1 kJ/mol and ΔS0= 669 J/mol K
A: Given ΔG° = -89.9 kJ/mol ΔH° = 83.1 kJ/mol ΔS° = 669 J/mol K
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: 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: For the reaction 2 A B, the value of Kc = 4.50, at 100oC. Determine the value of ∆G, in kJ/mol for…
A:
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: 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: Calculate the equilibrium constant of the reaction at 230.0 ∘C. Calculate the equilibrium constant…
A: Given thermodynamic data Substance ΔfH∘ (kJ mol−1) ΔfG∘ (kJ mol−1) S∘ (J mol−1 K−1)…
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: 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: The equilibrium constant for the reaction Na2(g) → 2Na(g) is 2.47 at 1000. K. Calculate the value of…
A: Equilibrium is the state at which the rate of forward and backward reaction are equal in a…
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: Le Chatelier's Principle: Calculate ΔH if K doubles when T increases from 298 K to 308 K. (answer =…
A: The Arrhenius equation is: logK2 K1=Ea2.303R1 T1-1 T2 Here K is the rate constant. Ea is the…
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)
A:
Q: For a certain chemical reaction, the equilibrium constant K = 2.2 x 10" at 5.00 °C. Calculate the…
A: The answer is given as follows -23 kJ is answer
Q: A reaction has an equilibrium constant of 8700 at 298 K. At 712 K, the equilibrium constant is 0.33.…
A: The formula used to calculate the enthalpy of reaction is,
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: For a spontaneous endothermic chemical reaction with ArxnH°= +20.6 kJ/mol, the equilibrium constant…
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 ∆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: For the reaction 2 A (g) → B (g), Kp = 0.00347 at 298 K. When ∆G = 8.35 kJ/mol, what is the partial…
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: Consider the following reaction, known as the Haber process for the production of ammonia: N2(g) +…
A: The reaction by which ammonia is formed from H2 and N2 is known as Haber process.
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 =…
Determine the equilibrium constant for a reaction at 200.0 K if ∆G° =-11.70 kJ/mol. (R = 8.314 J/mol ･ K)
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 1 images
