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The standard enthalpy of formation of NaCl(s) is fH° = -411.15 kJ mol-1. Write the reaction...

The standard enthalpy of formation of NaCl(s) is fH° = -411.15 kJ mol-1. Write the reaction to which the above standard enthalpy of formation refers. What would be the change in internal energy for this formation reaction at 298.15K and the standard pressure of 1.00 bar pressure?

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Answer #1

Given the standard enthalpy of formation for NaCl is

The formation reaction can be given as

Na(s) + 0.5Cl2(g) ---> NaCl(s)

Standard enthalpy of formaton is defined as amount of heat released during the formation of 1 mole of given compound.at 298.15K and 1 Bar standard tempreature and pressure respectively.

We Know that internal energy is dependent on tempreature and is defined as

where is change in tempreature.

SO Here  =0 as formation reaction is occure at same tempreatue 298.15 K

so change in internal energy is

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