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Question
Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below:
\[\ce{2 Zn (s) + O2 (g) -> 2 ZnO (s); ∆H = - 693.8 kJ mol^{-1}}\]
(i) The enthalpy of two moles of \[\ce{ZnO}\] is less than the total enthalpy of two moles of \[\ce{Zn}\] and one mole of oxygen by 693.8 kJ.
(ii) The enthalpy of two moles of \[\ce{ZnO}\] is more than the total enthalpy of two moles of \[\ce{Zn}\] and one mole of oxygen by 693.8 kJ.
(iii) 693.8 kJ mol–1 energy is evolved in the reaction.
(iv) 693.8 kJ mol–1 energy is absorbed in the reaction.
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Solution
(i) The enthalpy of two moles of \[\ce{ZnO}\] is less than the total enthalpy of two moles of \[\ce{Zn}\] and one mole of oxygen by 693.8 kJ.
(iii) 693.8 kJ mol–1 energy is evolved in the reaction.
Explanation:
As we know from thermodynamics,
`Δ_rH^o` = `isuma_iΔ_fH^⊙` (products) - `suma_ib_iΔ_fH^⊙` (reactants)
Since, the above reaction is an exothermic reaction, the enthalpy of reactant is more than the enthalpy of product.
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