answer:To calculate the molar heat capacity of carbon dioxide gas at constant volume, we need to use the formula: C_V (molar heat capacity) = (q/n) / (ΔT) where: q = heat energy added = 2.00 Joules n = moles of carbon dioxide = 1.00 mole ΔT = change in temperature = 398 K - 298 K = 100 K Now, plug these values into the formula: C_V = (2.00 J) / (1.00 mole) / (100 K) C_V = 0.02 J/mol·K Therefore, the molar heat capacity of carbon dioxide gas at constant volume is 0.02 J/mol·K., To solve this problem, we can use the definition of molar heat capacity at constant volume, which is given by: Cv = (ΔQ/n)/ΔT where ΔQ is the heat added to the system, n is the number of moles of the gas, and ΔT is the change in temperature. We are given that ΔQ = 2.00 Joules, n = 1.00 moles, and ΔT = 398 K - 298 K = 100 K. Plugging these values in, we get: Cv = (2.00 J/1.00 mol) / (100 K) = 0.020 J/(mol·K) Therefore, the molar heat capacity of carbon dioxide gas at constant volume is 0.020 J/(mol·K).,To calculate the molar heat capacity of carbon dioxide gas at constant volume (Cv), we can use the formula: q = n * Cv * ΔT where q is the heat energy added (in Joules), n is the number of moles, and ΔT is the change in temperature (in Kelvin). We are given the following information: - Initial temperature (T1) = 298 K - Final temperature (T2) = 398 K - Heat energy added (q) = 2.00 Joules - Number of moles (n) = 1.00 mole First, let's calculate the change in temperature (ΔT): ΔT = T2 - T1 = 398 K - 298 K = 100 K Now, we can rearrange the formula to solve for Cv: Cv = q / (n * ΔT) Plugging in the given values: Cv = 2.00 J / (1.00 mol * 100 K) = 0.02 J/(mol*K) So, the molar heat capacity of carbon dioxide gas at constant volume is 0.02 J/(mol*K).

answer:To calculate the entropy change for the combustion of methane gas at a constant temperature, we can use the formula ΔS° = Σn * S°(products) - Σm * S°(reactants), where n and m are the stoichiometric coefficients of the balanced chemical equation. The standard molar entropy values for O2, CO2, and H2O are as follows: S°(O2) = 205.14 J/(mol·K) S°(CO2) = 213.74 J/(mol·K) S°(H2O) = 188.84 J/(mol·K) (for gaseous state) Now, let's plug these values into the formula: ΔS° = 1 * S°(CO2) + 2 * S°(H2O) - 1 * S°(CH4) + 2 * S°(O2) ΔS° = 1 * 213.74 + 2 * 188.84 - 1 * 186.25 + 2 * 205.14 ΔS° = 591.32 - 596.53 ΔS° ≈ -5.21 J/(mol·K) Therefore, the entropy change for the combustion of methane gas at a constant temperature of 298 K and a pressure of 1 atm is approximately -5.21 J/(mol·K).,To calculate the entropy change for the combustion of methane gas, we need the standard molar entropies of all the reactants and products involved in the reaction. The balanced chemical equation is: CH4 (g) + 2O2 (g) -> CO2 (g) + 2H2O (g) We are given the standard molar entropy of CH4 as 186.25 J/(mol·K). We also need the standard molar entropies of O2, CO2, and H2O. The standard molar entropies for these substances are: O2 (g): 205.15 J/(mol·K) CO2 (g): 213.79 J/(mol·K) H2O (g): 188.84 J/(mol·K) Now, we can calculate the entropy change (ΔS) for the reaction using the following formula: ΔS = ΣnS(products) - ΣnS(reactants) where n is the stoichiometric coefficient and S is the standard molar entropy. ΔS = (1 × 213.79) + (2 × 188.84) - (1 × 186.25) + (2 × 205.15) ΔS = (213.79 + 377.68) - (186.25 + 410.30) ΔS = 591.47 - 596.55 ΔS = -5.08 J/(mol·K) The entropy change for the combustion of methane gas at a constant temperature of 298 K and a pressure of 1 atm is -5.08 J/(mol·K).,To calculate the entropy change for the combustion of methane gas, we first need to find the standard molar entropies of all the products and reactants involved in the reaction. The entropy change, ΔS, can then be calculated using the difference between the standard molar entropies of the products and reactants. From the balanced chemical equation: CH4 (g) + 2O2 (g) -> CO2 (g) + 2H2O (g) We know the standard molar entropy of CH4 is 186.25 J/(mol·K). To find the standard molar entropies of the products CO2 and H2O, we can consult a standard thermodynamic table, which gives the following values: ΔS_product(CO2) = -394.4 J/(mol·K) ΔS_product(H2O) = 69.95 J/(mol·K) Now, we can calculate the entropy change using the formula: ΔS = Σ ΔS_products - Σ ΔS_reactants ΔS = (-394.4) + (2 * 69.95) - 186.25 + (2 * 0) Note that the standard molar entropy of O2 is 0 J/(mol·K) under these conditions, as it is in its standard state. ΔS = (-394.4 + 139.9) - 186.25 ΔS = (-254.5) - 186.25 ΔS = -440.75 J/(mol·K) The entropy change for the combustion of methane gas at 298 K and 1 atm is -440.75 J/(mol·K).

answer:To calculate the Gibbs free energy change for the reaction, we first need to write the chemical equation for the reaction: H₂ (g) + Cl₂ (g) → 2HCl (g) Next, we need the standard Gibbs free energy of formation (ΔGf°) values for the elements and compounds involved. The ΔGf° values for the elements in their standard states are: H₂ (g) = 0 kJ/mol Cl₂ (g) = 0 kJ/mol HCl (g) = -92.3 kJ/mol Now, we can calculate the Gibbs free energy change for the reaction using the following equation: ΔG° = ΣΔGf°(products) - ΣΔGf°(reactants) ΔG° = (1 × -92.3) - (-92.3) - (-1 × 0) + (-1 × 0) ΔG° = 92.3 - 0 ΔG° = 92.3 kJ/mol The Gibbs free energy change for the reaction of hydrogen gas and chlorine gas to form hydrochloric acid at standard conditions (298 K and 1 atm) is 92.3 kJ/mol., To calculate the Gibbs free energy change for a reaction, we first need to find the standard Gibbs free energy change of formation (ΔG°f) for each substance involved in the reaction. Once we have these values, we can use them to calculate the overall standard Gibbs free energy change (ΔG°) for the reaction using the equation: ΔG° = ΣΔG°f(products) - ΣΔG°f(reactants) The standard Gibbs free energy change of formation is the change in Gibbs free energy when one mole of a substance is formed from its elements in their standard states at 1 atm and 298 K. For the reaction: H2(g) + Cl2(g) -> 2HCl(g) We need to find the ΔG°f values for H2(g), Cl2(g), and 2HCl(g). The ΔG°f for H2(g) and Cl2(g) are zero because they are elements in their standard states. The ΔG°f for HCl(g) is -95.3 kJ/mol. Now, we can plug these values into the equation to find ΔG°: ΔG° = ΔG°f(2HCl(g)) - ΔG°f(H2(g)) + ΔG°f(Cl2(g)) ΔG° = -95.3 kJ/mol * 2 - 0 + 0 ΔG° = -190.6 kJ/mol So, the Gibbs free energy change for the reaction of hydrogen gas and chlorine gas to form hydrochloric acid at standard conditions (298 K and 1 atm) is -190.6 kJ/mol. This indicates that the reaction is spontaneous and exothermic.,To calculate the Gibbs free energy change (ΔG) for the reaction of hydrogen gas (H2) and chlorine gas (Cl2) to form hydrochloric acid (2HCl) at standard conditions (298 K and 1 atm), we can use the following equation: ΔG = ΔH - TΔS where ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy. First, we need to find the standard enthalpy change (ΔH°) and standard entropy change (ΔS°) for the reaction. These values can be found in standard thermodynamic tables. For the given reaction: H2(g) + Cl2(g) → 2HCl(g) The standard enthalpy change (ΔH°) and standard entropy change (ΔS°) are as follows: ΔH°(H2) = 0 kJ/mol (since it is an element in its standard state) ΔH°(Cl2) = 0 kJ/mol (since it is an element in its standard state) ΔH°(HCl) = -92.3 kJ/mol ΔS°(H2) = 130.7 J/mol·K ΔS°(Cl2) = 223.1 J/mol·K ΔS°(HCl) = 186.8 J/mol·K Now, we can calculate the changes in enthalpy (ΔH) and entropy (ΔS) for the reaction: ΔH = 2 * ΔH°(HCl) - ΔH°(H2) + ΔH°(Cl2) = 2 * (-92.3) - (0 + 0) = -184.6 kJ/mol ΔS = 2 * ΔS°(HCl) - ΔS°(H2) + ΔS°(Cl2) = 2 * 186.8 - (130.7 + 223.1) = 373.6 - 353.8 = 19.8 J/mol·K Now, we can calculate the Gibbs free energy change (ΔG) at 298 K: ΔG = ΔH - TΔS = -184.6 kJ/mol - (298 K * 19.8 J/mol·K / 1000 J/kJ) = -184.6 kJ/mol - 5.9 kJ/mol = -190.5 kJ/mol So, the Gibbs free energy change for the reaction of hydrogen gas (H2) and chlorine gas (Cl2) to form hydrochloric acid (2HCl) at standard conditions (298 K and 1 atm) is -190.5 kJ/mol.

answer:To calculate the Gibbs free energy change (ΔG), we can use the equation: ΔG = ΔH - TΔS where: ΔG = Gibbs free energy change ΔH = Enthalpy change T = Temperature in Kelvin ΔS = Entropy change In this case, the temperature T is given as 298 K. Now, we can plug in the values: ΔG = (-483.6 kJ/mol) - (298 K * (-109.4 J/K/mol)) First, let's convert the units of ΔH and ΔS to the same unit (Joules). We know that 1 kJ = 1000 J, so: ΔH = -483.6 * 1000 = -483600 J/mol ΔS = -109.4 J/K/mol Now, we can plug these values back into the equation: ΔG = (-483600 J/mol) - (298 K * (-109.4 J/K/mol)) ΔG = (-483600 J/mol) - (-32239.2 J/mol) ΔG = -451360.8 J/mol To express the answer in kJ/mol instead of J/mol, divide the result by 1000: ΔG ≈ -451.4 kJ/mol So, the Gibbs free energy change for the given reaction at 298 K is approximately -451.4 kJ/mol.,To calculate the Gibbs free energy change (ΔG) for the reaction, we can use the equation: ΔG = ΔH - TΔS where ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy. Given, ΔH = -483.6 kJ/mol and ΔS = -109.4 J/K/mol. Then, we need to convert the entropy value from J/K/mol to J/K: -109.4 J/K/mol × (1000 J/1 kJ) = -109400 J/K/mol Now, we can plug in the values into the equation: ΔG = (-483.6 kJ/mol) - (298 K) × (-109400 J/K/mol) To convert ΔG from kJ/mol to J/K, we can multiply the result by (1 kJ/1000 J): ΔG = (-483.6 kJ/mol) - (298 K) × (-109400 J/K/mol) × (1 kJ/1000 J) ΔG = -0.183 kJ/mol Therefore, the Gibbs free energy change for the reaction at 298 K is -0.183 kJ/mol.,To calculate the Gibbs free energy change (ΔG) for the reaction, we can use the following equation: ΔG = ΔH - TΔS where ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy. Given values: ΔH = -483.6 kJ/mol ΔS = -109.4 J/K/mol T = 298 K First, we need to convert ΔS to kJ/K/mol: ΔS = -109.4 J/K/mol × (1 kJ / 1000 J) = -0.1094 kJ/K/mol Now, we can plug the values into the equation: ΔG = (-483.6 kJ/mol) - (298 K × -0.1094 kJ/K/mol) ΔG = -483.6 kJ/mol + 32.6 kJ/mol ΔG = -451 kJ/mol So, the Gibbs free energy change for the reaction is -451 kJ/mol.