Consider the reaction 2S(s) + 3O_2(g) rightarrow 2SO_3(g) Delta H degree = -791

Question

Consider the reaction 2S(s) + 3O_2(g) rightarrow 2SO_3(g) Delta H degree = -791 kJ Is the reaction endothermic or exothermic? Is it spontaneous? How do you know? Calculate the amount of heat evolved if 59.5 g of oxygen reacts with excess sulfur at constant pressure. Calculate the mass of SO_3(g) formed if 39.8 kJ of heat were released. Can you determine the quantity of heat is needed to decompose 132.5 g of SO_3(g) into its constituent elements? Can you use the reaction in a to calculate this heat? When the problem states "constituent elements", is there an assumption made about the state of those elements?

Explanation / Answer

since enthalpy change is -ve, the reaction is exothermic. That is heat is evolved during the course of reaction. For knowing whether the reaction is spontaneous, one has to use the following equation

deltaG ( Gibbs free energy change)= deltaH-T*deltaS

if deltaG is -ve, the Reaction is spontaneous. One has to Calculate the entropy change and substitute in the above equation, if deltaG is -ve, the reaction is spontaneous.

2. from the reaction, 2S(s)+ 3O2(g) ------->2SO3(g), molar mass of O2= 32 and that of SO3= 80

96 gm of O2 gives 791 KJ of heat

59.5 gm of O2 gives 59.5*791/96= 490.25 Kj of heat

3. 791 Kj of heat requires 2 moles of SO3 or 2*80= 160 gm of SO3

39.8 Kj of heat requires 39.8*160/791=8.05 gm of SO3

4. When SO3 decomposes heat is absorbed.

hence 160 gm of SO3 requires 791 Kj of heat to decompose

132.5 gm of SO3 requires 132.5*791/160= 655 Kj of heat has to be added.

The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions.

so this is nothing to do it with the state of the elements.

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