Tetraoxosulphate (vi) acid is prepared using the chemical reaction…
Tetraoxosulphate (vi) acid is prepared using the chemical reaction
SO3(g) + H2O(s) → H2SO4(l).
Given the heats of formation for SO3(g), H2O(l) and H2SO4(l) as -395KJmol-1, -286KJmol-1 and – 811KJmol-1 respectively, the enthalpy change accompanying this reaction is
- A. -1032KJ
- B. -130KJ
- C. +130KJ
- D. +1032KJ
Enthalpy change = heat of formation of product – heat of formation of reactants.
Heat of formation of product = heat of formation of sulphuric acid = – 811KJ/mol
Heat of formation of reactants = heat of formation of sulphur (vi) oxide + heat of formation of water = (-395) + (-286) = -395 – 286 = – 681KJ/mol
Hence, enthalpy change = (- 811) – (- 681) = – 811 + 681 = – 130 KJ/mol
Now for the right answer to the above question:
- Option A is incorrect.
- Option B is correct.
- C is incorrect.
- D is not the correct answer.
You may please note these/this:
- Always remember, heat of reaction or enthalpy change = heat of formation of product(s) – heat of formation of reactant(s)
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/ culled from 2017 JAMB-UTME Chemistry question 3 /
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