Consider the following reactions :

(i) H+(aq) + OH-(aq) = H2O(l) H = -x1kJ mol-1

(ii) H2(g) + 12O2(g) = H2O(l) H = -x2kJ mol-1

(iii) CO2(g) + H2(g) = CO(g) + H2O(l) H = -x3kJ mol-1

(iv) C2H5(g) + 52O2(g) = 2CO2(g) + H2O(l) H = -x4kJ mol-1

Enthalpy of formation of H2O(l) is:

(a) -x2 kJ mol-1 

(b) +x3 kJ mol-1

(c) -x4 kJ mol-1

(d) -x1 kJ mol-1

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Concept Questions :-

Enthalpy and It's Type

(a) Enthalpy of formation: The amount of heat evolved or absorbed during the formation of 1 mole of a compound from its constituent elements is known as the heat of formation. So, the correct answer is:

H2(g) + 12O2(g)           H2O(l), H=-x2kJmol-1

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