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Consider the following reaction $C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) ⟶ 6 {CO}_{2} (g) + 3 H_{2} O (g)$ signs of $Δ H, Δ S and Δ G$ for the above reaction will be

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+,+,_

-,+,+

+,-,+

-,+,-

answer is B.

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Detailed Solution

$\begin{array}{l} ΔH = - ve (Combustion reaction) \\ ΔS = + ve ({Δn}_{g} = + ve) \\ ΔG = ΔH - TΔS = - ve \end{array}$

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