Αποτελέσματα Αναζήτησης
Calculate the change in entropy of the surroundings when water freezes at −10.°C; use ΔHfus(H2O) = 6.0 kJ·mol−1 at −10.°C. Answer: We can expect the entropy of the surroundings to increase when water freezes because the heat released stirs up the thermal motion of the atoms in the surroundings.
Define Gibb’s free energy, enthalpy, and entropy, and the relationship between them. Predict the sign of ΔS in the following processes. Briefly explain your answer: Calculate ΔS°, ΔH°, and ΔG° for following reactions using standard values of formation.
Problem Set 12 Solutions. 1. What is the increase in entropy of one gram of ice at OoC is melted and heated to 500C? The change in entropy is given by dS = dQ. T . In this case, the dQ must be calculated in two pieces. First there is the heat needed to melt the ice, and then there is the heat needed to raise the temperature of the system.
When 1.5 kg of an ideal gas (specific heat at constant volume 0.8216 kJ kg K ) is heated = × at constant volume to a final temperature of 425°C, the total entropy increase is 0.4386 kJ/K. The initial temperature of the gas is most nearly. 200°C. 210°C.
Global changes in entropy. We know that reactions like the formation of ammonia from nitrogen and hydrogen actually do happen despite the decrease in entropy of the system. What is the reason? Calculate the change in entropy of the surroundings when water freezes at −10.°C; use ΔHfus(H2O) = 6.0 kJ·mol−1 at −10.°C.
Calculate the change in the entropies of the system and the surroundings, and the total entropy change when a sample of oxygen gas of mass 50 g at 300 K and 1 bar doubles its volume in:
1. Calculate the change in entropy of a large vat of molten copper when 50 J of energy is removed reversible from it as heat at 1100 °C. 2. Calculate the change in entropy of 1.0 L of water at 0 °C when it absorbs 235 J of energy from a heater.
