The standard molar entropy of lead(II) bromide (PbBr2) is 161 J/mol K. What is t

Question

The standard molar entropy of lead(II) bromide (PbBr2) is 161 J/mol K. What is the entropy of What is the entropy change in the surroundings when one mole of ice melts at 0.00°C in a large room maintained at 27.0°C? The heat of fusion of ice is 6.31 kJ/mol. (answer in J/K) The enthalpy of vaporization of water at 100.0°C is 40.55 kJ/mol. What is the entropy change in the surroundings when one mole of water vapor condenses at 100.0°C in a large room maintained at a temperature of 15.00°C? (answer in J/K) The enthalpy and entropy of fusion of ice at 0.00°C are 6.00 kJ/mol and 21.5 J/mol K respectively. What is the entropy change of the universe when one mole of ice melts in a large room maintained at 30.00°C? Assume the final temperature of the water is 0.00°C. (answer in J/K) Determine AS for the reaction to form 45.92 g of, Zn(s)+ 2 HCKa)-ZnCl(ag)+Hg) given (answer in J) Substance S° (J/mol K) Zn(s) HCl(aq) 56.5 60.9 130.58 55.10 Zn (aq) 106.5 CI (aq) In an experiment, 1.000 atm of H2(g) in a 5.00 L container at 25.0°C was reacted under standard state conditions with a stoichiometric quantity of O+(g) to form water vapor. Determine the entropy change.(answer in J/K) Substance S(J/mol-K) H2(g) 0:(g) H20(g) 188.83 H200 69.95 130.58 205.0 Determine the standard entropy change of the universe when 0.4 mol of NHNO, is produced, NH4 (aq) +NO; (aq)- NH,NO,(s), given the following information. Is the reaction spontaneous under standard conditions? (answer in J/K) Substance },OkJ/mol) So(J/mol-K) NH4NO3(s) -365.6 NH4 (aq) 132.5 NO3 (aq) -205.0 151.1 113.4 146.4 Hydrogen reacts with nitrogen to form ammonia (NHs) according to the reaction . K. Determine Go at 50.00°C for the preparation of 2 moles of NH3. (answer in kumol 3112(g) + Ndg) 2 NHdg). The value of AIP is-92.38 kJrnol, and that of ASS-198.2 Jmol

Explanation / Answer

Given

standard molar entropy = 161 J/mol

Mass of PbBr2 = 2.45 g

Molar mass of  PbBr2 = 367 g/mol

No. of moles of PbBr2 = 2.45 g / 367 g/mol = 6.676 * 10-3 moles

Entropy = No. of moles * standard entropy =  6.676 * 10-3 moles * 161 J/mol = 1.075 J

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Given

temperature of ice = 0 C = 273 K

Temperature of room = 27 C = 300 K

differrence = 300 K -273 K = 27 K

Entropy = No. of moles * heat of fusion = 1 moles * 6.31 kJ/mol / (27- 0) = 233.7 J/K

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