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What is Δngas for the combustion of 1 mole of benzene, when both the reactants and the products are at 298 K?

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What is Δngas for the combustion of 1 mole of benzene, when both the reactants and the products are at 298 K?

(a) 0

(b) 1/2

(c) 3/2

(d) -3/2

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Correct answer: (d) -3/2

Explanation:

C6H6(l) + \(\frac{15}{2}\)O2(g) → 6CO2(g) + 3H2O(l)

Δn = 6 - \(\frac{15}{2}\)

= - \(\frac{3}{2}\)

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