What is the amount of work done when 0.5 mol | vedclass

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(D) The combustion reaction is: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$.Work done is given by the formula: $w = -\Delta n_g RT$.Fo

Vedclass · Accepted Answer

$+2494 \ J$

Vedclass · Answer

(D) The combustion reaction is: $CH_{4(g)} + 2O_{2(g)} ightarrow CO_{2(g)} + 2H_2O_{(l)}$.Work done is given by the formula: $w = -\Delta n_g RT$.For $1 \ mol$ of $CH_4$,the change in gaseous moles is $\Delta n_g = n_{g(products)} - n_{g(reactants)} = 1 - (1 + 2) = -2$.For $0.5 \ mol$ of $CH_4$,$\Delta n_g = 0.5 	imes (-2) = -1 \ mol$.Substituting the values: $w = -(-1 \ mol) 	imes (8.314 \ J \ K^{-1} \ mol^{-1}) 	imes (300 \ K)$.$w = +2494 \ J$.

What is the amount of work done when $0.5 \ mol$ of methane,$CH_{4(g)}$,is subjected to combustion at $300 \ K$? (Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

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