Consider the reaction P4010(s) + 6H₂O(1) 4H3PO4(aq) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.57 moles a P4010(s) react at standard conditions. AS surroundings = J/K

Consider the reaction 2NH3(g) + 3N₂O(g) → 4N₂ (9) + 3H₂O(g) for which AH° = − 879.5 kJ and AS° = 288.1 J/K at 298.15 K. 1. Calculate the entropy change of the UNIVERSE when 2.344 moles of NH3(g) react under standard conditions at 298.15 K. ASuniverse = J/K 2. Is this reaction reactant or product favored under standard conditions? 3. If the reaction is product favored, is it enthalpy favored, entropy favored, or favored by both enthalpy and entropy? If the reaction is reactant favored, choose 'reactant favored'.

A chemical engineer is studying the two reactions shown in the table below. In each case, she fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 61.0 °C and constant total pressure. Then, she measures the reaction enthalpy AH and reaction entropy as of the first reaction, and the reaction enthalpy H and reaction free energy AG of the second reaction. The results of her measurements are shown in the table. Complete the table. That is, calculate AG for the first reaction and As for the second. (Round your answer to zero decimal places.) Then, decide whether, under the conditions the engineer has set up, the reaction is spontaneous, the reverse reaction is spontaneous, or neither forward nor reverse reaction is spontaneous because the system is at equilibrium. 4H₂O₂(1) + PbS (s) PbSO₂ (s) + 4H₂0 (1) Sn (s) + 2CO₂ (g) SnO₂ (s) + 2CO (g) ΔΗ = - 1212. kJ AS = -3559. AG = KJ Which is spontaneous? Othis reaction O the reverse reaction O neither…

Calculate the change in entropy for a sample heated from 298 K to 500 K under constant pressure. The constant pressure heat capacity is 29 J/mol K.