Question

Calculate Δ G ∘ rxn for the following reaction: 4CO(g)+2N O 2 (g)→4C O 2 (g)+ N 2 (g) . Use the following reactions and given Δ G ∘ rxn values: 2NO(g)+ O 2 (g)→2N O 2 (g) , Δ G ∘ rxn = - 72.6 kJ 2CO(g)+ O 2 (g)→2C O 2 (g) , Δ G ∘ rxn = - 514.4 kJ 1 2 O 2 (g)+ 1 2 N 2 (g)→NO(g) , Δ G ∘ rxn = 87.6 kJ Express your answer using one decimal place. Δ G ∘ rxn = nothing kJ 2) Consider the following reaction:
CO(g)+2H2(g)⇌CH3OH(g)
Kp=2.26×104 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under each of the following conditions.​   

Part A

standard conditions

ΔGrxn =		kJ

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Part B

at equilibrium

ΔGrxn =		kJ

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Part C

PCH3OH= 1.1 atm ;
PCO=PH2= 1.5×10⁻² atm

Express your answer using two significant figures.

3)

For the decomposition of barium carbonate, consider the following thermodynamic data (Due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.): ΔH∘rxn 243.5kJ/mol ΔS∘rxn 172.0J/(mol⋅K) Calculate the temperature in kelvins above which this reaction is spontaneous. Express your answer to four significant figures and include the appropriate units. View Available Hint(s) SubmitPrevious AnswersRequest Answer Relationship between free energy and the equilibrium constant The standard free energy change, ΔG∘ , and the equilibrium constant K for a reaction can be related by the following equation: ΔG∘=−RTlnK where T is the Kelvin temperature and R is equal to 8.314 J/(mol⋅K) . Part B Part complete The thermodynamic values from part A will be useful as you work through part B: T = nothing nothing ΔH∘rxn 243.5kJ/mol ΔS∘rxn 172.0J/(mol⋅K) Calculate the equilibrium constant for the following reaction at room temperature, 25 ∘C : BaCO3(s)→BaO(s)+CO2(g) K = nothing

Answer 1