answer:To calculate the standard enthalpy change for the fusion of 50 grams of ice at -10°C to liquid water at 10°C, we need to consider three steps: 1. Heating the ice from -10°C to 0°C. 2. Melting the ice at 0°C. 3. Heating the liquid water from 0°C to 10°C. First, let's calculate the moles of ice: The molar mass of water (H2O) is 18.015 g/mol. moles = (mass) / (molar mass) = 50 g / 18.015 g/mol = 2.776 moles Step 1: Heating the ice from -10°C to 0°C The specific heat capacity of ice is approximately 2.093 J/g°C. q1 = (mass) × (specific heat capacity of ice) × (change in temperature) q1 = 50 g × 2.093 J/g°C × (0 - (-10))°C = 50 g × 2.093 J/g°C × 10°C = 1046.5 J Step 2: Melting the ice at 0°C q2 = (moles) × (enthalpy of fusion) q2 = 2.776 moles × 6.01 kJ/mol = 16.681 kJ = 16681 J Step 3: Heating the liquid water from 0°C to 10°C q3 = (mass) × (specific heat capacity of water) × (change in temperature) q3 = 50 g × 4.184 J/g°C × (10 - 0)°C = 50 g × 4.184 J/g°C × 10°C = 2092 J Now, let's sum up the enthalpy changes for all three steps: ΔH = q1 + q2 + q3 = 1046.5 J + 16681 J + 2092 J = 19819.5 J The standard enthalpy change for the fusion of 50 grams of ice at -10°C to liquid water at 10°C is 19.82 kJ (rounded to two decimal places).

answer:The standard enthalpy change for the fusion reaction of solid sulfur at its melting point to become liquid sulfur at its boiling point can be calculated by considering two separate processes: the melting of solid sulfur (fusion) and the heating of liquid sulfur from its melting point to its boiling point. 1. Enthalpy change of fusion (ΔHfus): The enthalpy change when one mole of a solid substance melts at its melting point under standard conditions (1 atm pressure). For sulfur, the standard enthalpy of fusion is 1.72 kJ/mol. 2. Enthalpy change of heating (ΔHheat): The enthalpy change when one mole of a substance is heated from its melting point to its boiling point. This can be calculated using the formula ΔHheat = m × C × ΔT, where m is the number of moles, C is the specific heat capacity of the substance, and ΔT is the temperature change. For sulfur, the specific heat capacity of liquid sulfur is approximately 1.09 J/g·K. The melting point of sulfur is 115.21°C, and its boiling point is 444.6°C. The molar mass of sulfur is 32.06 g/mol. First, convert the temperature change from Celsius to Kelvin: ΔT = (444.6 - 115.21) K = 329.39 K. Next, calculate the number of grams in one mole of sulfur: 1 mol × 32.06 g/mol = 32.06 g. Now, calculate the enthalpy change of heating: ΔHheat = (32.06 g) × (1.09 J/g·K) × (329.39 K) = 1150.8 J = 1.15 kJ (rounded to two decimal places). Finally, add the enthalpy change of fusion and the enthalpy change of heating to find the total standard enthalpy change for the process: ΔHtotal = ΔHfus + ΔHheat = 1.72 kJ + 1.15 kJ = 2.87 kJ/mol.

answer:To calculate the standard enthalpy change for the fusion of 2.5 moles of solid sulfur (S8) at its melting point (388 K), we can use the given information: ΔHfus = 9.73 kJ/mol (enthalpy of fusion) ΔSfus = 22.8 J/(mol*K) (entropy of fusion) First, we need to find the enthalpy change for the fusion of 2.5 moles of sulfur. We can do this by multiplying the enthalpy of fusion per mole by the number of moles: ΔH = ΔHfus * moles ΔH = 9.73 kJ/mol * 2.5 mol ΔH = 24.325 kJ Now, we have the enthalpy change for the fusion of 2.5 moles of sulfur. However, we also need to account for the heat capacity change during the phase transition. We can do this by using the heat capacities of solid and liquid sulfur and the temperature at which the phase transition occurs: ΔCp = Cp(liquid) - Cp(solid) ΔCp = 37.05 J/(mol*K) - 22.75 J/(mol*K) ΔCp = 14.3 J/(mol*K) Now, we can calculate the additional enthalpy change due to the heat capacity change: ΔH_additional = ΔCp * moles * ΔT ΔH_additional = 14.3 J/(mol*K) * 2.5 mol * (388 K - 388 K) ΔH_additional = 0 kJ Since the temperature remains constant during the phase transition, there is no additional enthalpy change due to the heat capacity change. Finally, we can find the total standard enthalpy change for the fusion of 2.5 moles of solid sulfur: ΔH_total = ΔH + ΔH_additional ΔH_total = 24.325 kJ + 0 kJ ΔH_total = 24.325 kJ The standard enthalpy change for the fusion of 2.5 moles of solid sulfur (S8) at its melting point (388 K) is 24.325 kJ.

answer:The standard enthalpy change for the fusion of solid hydrogen (H₂) at its melting point can be calculated using the given standard enthalpy of fusion. In this case, the standard enthalpy of fusion is already provided as 0.117 kJ/mol. Therefore, the standard enthalpy change for the fusion of solid hydrogen (H₂) at its melting point of 13.81 K is 0.117 kJ/mol.