For the reaction N2(g) + 3H2(g) <=> 2NH3(g) with ΔH = -92 kJ/mol, state the equilibrium law and the effect of changes in concentration, temperature, and pressure on the equilibrium position.

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Asked on May 26, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-05-26T23:02:37.598Z

Welcome back jolivettemequathotsea — been a while! Let's pick up where you left off. Parts (A) and (B) of the question require experimental data (e.g., initial rates at different concentrations) to determine the order of reaction and calculate the rate constant. This data is not provided in the image. Therefore, these parts cannot be solved. c) i) State the equilibrium law. The equilibrium law expresses the relationship between the concentrations (or partial pressures) of products and reactants at equilibrium. For the given reaction: N_2(g) + 3H_2(g) 2NH_3(g) The equilibrium constant expression in terms of concentrations (K_c) is: K_c = [NH_3]^2[N_2][H_2]^3 The equilibrium constant expression in terms of partial pressures (K_p) is: K_p = (P_NH_3)^2(P_N_2)(P_H_2)^3 The equilibrium law states that at a given temperature, the ratio of the product of the concentrations (or partial pressures) of the products raised to their stoichiometric coefficients to the product of the concentrations (or partial pressures) of the reactants raised to their stoichiometric coefficients is a constant. ii) State the effect of the following on the equilibrium position for the reaction above: The reaction is N_2(g) + 3H_2(g) 2NH_3(g) with H = -92 kJ/mol. This indicates the forward reaction is exothermic. (A) Increase in temperature: Step 1: Identify the nature of the reaction's enthalpy change. The forward reaction is exothermic ( H = -92 kJ/mol). Step 2: Apply Le Chatelier's principle. An increase in temperature will shift the equilibrium in the endothermic direction to absorb the added heat. The reverse reaction (decomposition of NH_3) is endothermic. The equilibrium position will shift to the left, favoring the formation of reactants (N_2 and H_2). (B) Decrease in pressure: Step 1: Determine the number of moles of gaseous reactants and products. Reactant side: 1 mol N_2 + 3 mol H_2 = 4 moles of gas Product side: 2 mol NH_3 = 2 moles of gas Step 2: Apply Le Chatelier's principle. A decrease in pressure will shift the equilibrium to the side with a greater number of moles of gas to counteract the pressure change. The equilibrium position will shift to the left, favoring the formation of reactants (N_2 and H_2). d) Given: Vapor pressure of pure ethanol (P^_ethanol) = 44 mmHg Vapor pressure of pure methanol (P^_methanol) = 94 mmHg Mass of methanol (CH_3OH) = 30g Mass of ethanol (CH_3CH_2OH) = 45g RAM: C=12, O=16, H=1 i) Calculate the mole fraction of each component in the solution. Step 1: Calculate the molar mass of each component. Molar mass of CH_3OH = (1 × 12) + (4 × 1) + (1 × 16) = 12 + 4 + 16 = 32 g/mol Molar mass of CH_3CH_2OH = (2 × 12) + (6 × 1) + (1 × 16) = 24 + 6 + 16 = 46 g/mol Step 2: Calculate the moles of each component. Moles of methanol (n_methanol): n_methanol = 30 g32 g/mol = 0.9375 mol Moles of ethanol (n_ethanol): n_ethanol = 45 g46 g/mol ≈ 0.9783 mol Step 3: Calculate the total moles in the solution. n_total = n_methanol + n_ethanol = 0.9375 mol + 0.9783 mol = 1.9158 mol Step 4: Calculate the mole fraction of each component. Mole fraction of methanol (X_methanol): X_methanol = n_methanoln_total = 0.9375 mol1.9158 mol ≈ 0.4894 Mole fraction of ethanol (X_ethanol): X_ethanol = n_ethanoln_total = 0.9783 mol1.9158 mol ≈ 0.5106 ii) Calculate the partial vapor pressure of each component in the solution at 20°C. Step 1: Apply Raoult's Law (P_i = X_i P^_i). Partial vapor pressure of methanol (P_methanol): P_methanol = X_methanol × P^_methanol = 0.4894 × 94 mmHg ≈ 46.00 mmHg Partial vapor pressure of ethanol (P_ethanol): P_ethanol = X_ethanol × P^_ethanol = 0.5106 × 44 mmHg ≈ 22.47 mmHg iii) Calculate the total vapor pressure of the mixture. Step 1: Apply Dalton's Law of Partial Pressures (P_total = P_methanol + P_ethanol). P_total = 46.00 mmHg + 22.47 mmHg = 68.47 mmHg 2. (a) (i) Define the term standard enthalpy of formation. The *standard enthalpy

Accepted Answer · 2026-05-26T23:02:37.598Z

Welcome back jolivettemequathotsea — been a while! Let's pick up where you left off. Parts (A) and (B) of the question require experimental data (e.g., initial rates at different concentrations) to determine the order of reaction and calculate the rate constant. This data is not provided in the image. Therefore, these parts cannot be solved. c) i) State the equilibrium law. The equilibrium law expresses the relationship between the concentrations (or partial pressures) of products and reactants at equilibrium. For the given reaction: N_2(g) + 3H_2(g) 2NH_3(g) The equilibrium constant expression in terms of concentrations (K_c) is: K_c = [NH_3]^2[N_2][H_2]^3 The equilibrium constant expression in terms of partial pressures (K_p) is: K_p = (P_NH_3)^2(P_N_2)(P_H_2)^3 The equilibrium law states that at a given temperature, the ratio of the product of the concentrations (or partial pressures) of the products raised to their stoichiometric coefficients to the product of the concentrations (or partial pressures) of the reactants raised to their stoichiometric coefficients is a constant. ii) State the effect of the following on the equilibrium position for the reaction above: The reaction is N_2(g) + 3H_2(g) 2NH_3(g) with H = -92 kJ/mol. This indicates the forward reaction is exothermic. (A) Increase in temperature: Step 1: Identify the nature of the reaction's enthalpy change. The forward reaction is exothermic ( H = -92 kJ/mol). Step 2: Apply Le Chatelier's principle. An increase in temperature will shift the equilibrium in the endothermic direction to absorb the added heat. The reverse reaction (decomposition of NH_3) is endothermic. The equilibrium position will shift to the left, favoring the formation of reactants (N_2 and H_2). (B) Decrease in pressure: Step 1: Determine the number of moles of gaseous reactants and products. Reactant side: 1 mol N_2 + 3 mol H_2 = 4 moles of gas Product side: 2 mol NH_3 = 2 moles of gas Step 2: Apply Le Chatelier's principle. A decrease in pressure will shift the equilibrium to the side with a greater number of moles of gas to counteract the pressure change. The equilibrium position will shift to the left, favoring the formation of reactants (N_2 and H_2). d) Given: Vapor pressure of pure ethanol (P^_ethanol) = 44 mmHg Vapor pressure of pure methanol (P^_methanol) = 94 mmHg Mass of methanol (CH_3OH) = 30g Mass of ethanol (CH_3CH_2OH) = 45g RAM: C=12, O=16, H=1 i) Calculate the mole fraction of each component in the solution. Step 1: Calculate the molar mass of each component. Molar mass of CH_3OH = (1 × 12) + (4 × 1) + (1 × 16) = 12 + 4 + 16 = 32 g/mol Molar mass of CH_3CH_2OH = (2 × 12) + (6 × 1) + (1 × 16) = 24 + 6 + 16 = 46 g/mol Step 2: Calculate the moles of each component. Moles of methanol (n_methanol): n_methanol = 30 g32 g/mol = 0.9375 mol Moles of ethanol (n_ethanol): n_ethanol = 45 g46 g/mol ≈ 0.9783 mol Step 3: Calculate the total moles in the solution. n_total = n_methanol + n_ethanol = 0.9375 mol + 0.9783 mol = 1.9158 mol Step 4: Calculate the mole fraction of each component. Mole fraction of methanol (X_methanol): X_methanol = n_methanoln_total = 0.9375 mol1.9158 mol ≈ 0.4894 Mole fraction of ethanol (X_ethanol): X_ethanol = n_ethanoln_total = 0.9783 mol1.9158 mol ≈ 0.5106 ii) Calculate the partial vapor pressure of each component in the solution at 20°C. Step 1: Apply Raoult's Law (P_i = X_i P^_i). Partial vapor pressure of methanol (P_methanol): P_methanol = X_methanol × P^_methanol = 0.4894 × 94 mmHg ≈ 46.00 mmHg Partial vapor pressure of ethanol (P_ethanol): P_ethanol = X_ethanol × P^_ethanol = 0.5106 × 44 mmHg ≈ 22.47 mmHg iii) Calculate the total vapor pressure of the mixture. Step 1: Apply Dalton's Law of Partial Pressures (P_total = P_methanol + P_ethanol). P_total = 46.00 mmHg + 22.47 mmHg = 68.47 mmHg 2. (a) (i) Define the term standard enthalpy of formation. The *standard enthalpy

For the reaction N2(g) + 3H2(g) <=> 2NH3(g) with ΔH = -92 kJ/mol, state the equilibrium law and the effect of changes in concentration, temperature, and pressure on the equilibrium position.

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For the reaction N2(g) + 3H2(g) <=> 2NH3(g) with ΔH = -92 kJ/mol, state the equilibrium law and the effect of changes in concentration, temperature, and pressure on the equilibrium position.

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