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(B) The reaction is $4 HCl_{(g)} + O_{2_{(g)}} \rightarrow 2 Cl_{2_{(g)}} + 2 H_{2}O_{(g)}$.First,calculate the change in the number of moles of gaseo

Vedclass · Accepted Answer

$623.6$

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(B) The reaction is $4 HCl_{(g)} + O_{2_{(g)}} ightarrow 2 Cl_{2_{(g)}} + 2 H_{2}O_{(g)}$.First,calculate the change in the number of moles of gaseous species,$\Delta n_g = (n_{products} - n_{reactants}) = (2 + 2) - (4 + 1) = 4 - 5 = -1$.This $\Delta n_g$ is for $4 \ moles$ of $HCl$.For $1 \ mole$ of $HCl$,$\Delta n_g = -1 / 4 = -0.25$.The work done is given by $W = -\Delta n_g RT$.Given $T = 27^{\circ} C = 300 \ K$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.$W = -(-0.25) 	imes 8.314 	imes 300$.$W = 0.25 	imes 2494.2 = 623.55 \ J \approx 623.6 \ J$.

Calculate the work done in the oxidation of one mole $HCl_{(g)}$ at $27^{\circ} C$,according to the reaction: $4 HCl_{(g)} + O_{2_{(g)}} \rightarrow 2 Cl_{2_{(g)}} + 2 H_{2}O_{(g)}$ $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$ (in $J$)

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