NEET Chemistry Enthalpy Match the Following Question
Match Column-I with Column-II.
Column-I Column-II (a) Standard Enthalpy Change (i) (sum of enthalpies of products) – (sum of enthalpies of reactants) (b) Enthalpy of Combustion (ii) -2.48 × 10² kJ mol⁻¹ (c) Enthalpy of Vaporization (iii) 44.01 kJ mol⁻¹ (d) Enthalpy of Fusion (iv) 6.00 kJ mol⁻¹
Choose the correct matching:
a-i, b-ii, c-iii, d-iv
a-ii, b-i, c-iv, d-iii
a-iv, b-iii, c-i, d-ii
a-iii, b-iv, c-ii, d-i
Correct Answer
Detailed Explanation
The standard enthalpy change is calculated as the difference between the sum of enthalpies of products and reactants (i). The enthalpy of combustion for graphite is given as -2.48 × 10² kJ mol⁻¹ (ii). The enthalpy of vaporization of water is 44.01 kJ mol⁻¹ (iii), and the enthalpy of fusion of ice is 6.00 kJ mol⁻¹ (iv).
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