Question

The enthalpy of formation for C6 H6 (I) is 49.0 kJ/mol. Consider the following reaction.

6 upper C (s, graphite) plus 3 upper H subscript 2 (g) right arrow upper C subscript 6 upper H subscript g (l).

Is the reaction endothermic or exothermic, and what is the enthalpy of reaction?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..
exothermic; Delta.Hrxn = 49.0 kJ
exothermic; Delta.Hrxn = –49.0 kJ
endothermic; Delta.Hrxn = 49.0 kJ
endothermic; Delta.Hrxn = –49.0 kJ

Answer 1

endothermic; ΔHrxn = 49.0 kJ

Reason: ΔHrxn = ΣΔHf(products) − ΣΔHf(reactants) = 49.0 kJ − 0 = 49.0 kJ (elements in standard states have ΔHf = 0), so the reaction absorbs heat.