What is the work done during the oxidation of | vedclass

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(NONE) The chemical equation for the oxidation of $4 \ mol$ of $SO_{2(g)}$ is: $4 SO_{2(g)} + 2 O_{2(g)} \longrightarrow 4 SO_{3(g)}$ Calculate the ch

Vedclass · Accepted Answer

$-4.988$

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(NONE) The chemical equation for the oxidation of $4 \ mol$ of $SO_{2(g)}$ is: $4 SO_{2(g)} + 2 O_{2(g)} \longrightarrow 4 SO_{3(g)}$ Calculate the change in the number of moles of gaseous species,$\Delta n_g$: $\Delta n_g = (n_{product, gas}) - (n_{reactant, gas}) = 4 - (4 + 2) = -2 \ mol$ The work done $(W)$ is given by the formula: $W = -\Delta n_g RT$ Given $T = 27^{\circ} C = 300 \ K$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$: $W = -(-2 \ mol) 	imes 8.314 \ J \ K^{-1} \ mol^{-1} 	imes 300 \ K$ $W = +4988.4 \ J = +4.988 \ kJ$ Since the calculated value is $+4.988 \ kJ$ and it does not match any of the given options,the provided options are incorrect.

What is the work done during the oxidation of $4 \ mol$ of $SO_{2(g)}$ to $SO_{3(g)}$ at $27^{\circ} C$ (in $kJ$)? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

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