Soal Termokimia

The given answer is the correct balanced chemical equation for the formation of crystalline Na2SO4. It shows that 2 moles of solid Na reacts with 1/8 mole of S8 and 2 moles of O2 to produce 1 mole of crystalline Na2SO4. This equation represents the formation of Na2SO4 from its constituent elements and is consistent with the law of conservation of mass.

The given reaction equation represents the combustion of glucose, where glucose reacts with oxygen to form carbon dioxide and water. The enthalpy change (∆H) for this reaction is -280 kJ, indicating that it is an exothermic reaction, releasing energy.

The second reaction equation represents the fermentation of glucose, where glucose is converted into ethanol and carbon dioxide. The enthalpy change (∆H) for this reaction is not given, but it can be determined by using the enthalpy change of combustion of ethanol.

The enthalpy change of combustion of ethanol is -1380 kJ, which means that the combustion of ethanol releases 1380 kJ of energy. Since 2 moles of ethanol are produced in the fermentation of 1 mole of glucose, the enthalpy change for the fermentation of glucose can be calculated as 2 times -1380 kJ, which is -2760 kJ.

Since the fermentation of glucose is an exothermic reaction, it releases energy. Therefore, the enthalpy change for the fermentation of glucose is -2760 kJ. However, the question asks for the enthalpy change in the opposite direction, which is the enthalpy change for the reaction of 2 moles of ethanol and 2 moles of carbon dioxide producing 1 mole of glucose.

Since the enthalpy change for the reverse reaction is the negative of the enthalpy change for the forward reaction, the enthalpy change for the reverse reaction is +2760 kJ. Therefore, the correct answer is +2880 kJ, which is the closest option to +2760 kJ.

The given chemical equation represents a reaction where 4 moles of ammonia (NH3) and 7 moles of oxygen (O2) react to form 4 moles of nitrogen dioxide (NO2) and 6 moles of water (H2O). The enthalpy change (∆H) for this reaction is -4 kJ. The question asks for the enthalpy change for the formation of NO2.

To find the enthalpy change for the formation of NO2, we need to consider the stoichiometric coefficients in the balanced equation. The stoichiometric coefficient for NO2 is 4, which means that 4 moles of NO2 are formed in the reaction.

The enthalpy change for the formation of NO2 can be calculated by multiplying the enthalpy change for the overall reaction by the ratio of the stoichiometric coefficient of NO2 to the stoichiometric coefficient of NH3. In this case, the ratio is (4/4) = 1.

Therefore, the enthalpy change for the formation of NO2 is equal to the enthalpy change for the overall reaction, which is (1 1/2a - b - c) kJ/mol.

The enthalpy change for the reaction CaO (s) + H2O (l) → Ca(OH)2 (s) is -64 kJ/mol. Since the enthalpy change for the reaction 2Ca (s) + O2(g) → 2CaO (s) is -1269 kJ, we can use this information to calculate the enthalpy change for the formation of Ca(OH)2. By dividing the enthalpy change of the formation reaction by the stoichiometric coefficient of CaO (s) in the reaction 2Ca (s) + O2(g) → 2CaO (s), we get -64 kJ/mol / 2 = -32 kJ/mol. Then, by adding this value to the enthalpy change of the reaction 2H2 (g) + O2 (g) → 2H2O (l) (-571 kJ/mol), we get -571 kJ/mol - 32 kJ/mol = -603 kJ/mol. However, since the reaction given is reversed, the enthalpy change should be positive, so the correct answer is -984 kJ/mol.

The given reaction is the combustion of methane (CH4) in the presence of oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The enthalpy change for the combustion of methane is the sum of the enthalpy changes for the individual reactions involved. The enthalpy change for the combustion of methane can be calculated by adding the enthalpy changes for the formation of CO2 and H2O and subtracting the enthalpy change for the formation of CH4. The enthalpy change for the formation of CO2 is -393.7 kJ/mol, the enthalpy change for the formation of H2O is -285.9 kJ/mol, and the enthalpy change for the formation of CH4 is -74.9 kJ/mol. Therefore, the enthalpy change for the combustion of methane is (-393.7 + 2*(-285.9)) - (-74.9) = -890.6 kJ.

The given question provides the enthalpy of formation (∆Hf) for H2O, CO2, and C2H2. The balanced equation for the combustion of C2H2 shows that 2 moles of C2H2 produce 4 moles of CO2 and 2 moles of H2O. Therefore, the combustion of 52 grams of C2H2 would produce (4/2) * (52/26) = 4 moles of CO2 and (2/2) * (52/26) = 2 moles of H2O. The total enthalpy change for the combustion reaction can be calculated by summing the products and subtracting the reactants: (4 * ∆Hf CO2) + (2 * ∆Hf H2O) - (2 * ∆Hf C2H2) = (4 * -394) + (2 * -242) - (2 * 52) = -1576 - 484 - 104 = -2164 kJ. Therefore, the correct answer is 2164 kJ.

The given reaction equation shows that the formation of 2 moles of CO2 releases 569 kJ of energy. Therefore, the formation of 1 mole of CO2 releases 569/2 = 284.5 kJ of energy. Since the question asks for the formation of 40 grams of carbon monoxide (which has a molar mass of 28 g/mol), we can calculate the number of moles of CO formed as 40/28 = 1.43 moles. Therefore, the energy released in the formation of 40 grams of CO is 1.43 * 284.5 = 407.4 kJ, which is approximately +175 kJ.

The given question states that the process of breaking down H2O into its atoms requires 220 kkal/mol of energy. Since breaking down a molecule requires energy, the energy needed to break the O-H bond in H2O is positive. Therefore, the average bond energy of O-H is +110 kkal/mol.

The correct answer is C (s) + O2 (g) → CO2 (g). This reaction involves the formation of carbon dioxide (CO2) from carbon (C) and oxygen (O2) in their standard states. The enthalpy change associated with this reaction is known as the heat of formation or the heat of combustion of carbon dioxide. It represents the amount of energy released or absorbed when one mole of carbon dioxide is formed from its elements in their standard states.