The work done during the combustion of 9 time | vedclass

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(C) The combustion reaction for ethane is: $C_2H_{6(g)} + \frac{7}{2}O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$.Change in gaseous moles,$\Delta n_

Vedclass · Accepted Answer

$18.71$

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(C) The combustion reaction for ethane is: $C_2H_{6(g)} + \frac{7}{2}O_{2(g)} ightarrow 2CO_{2(g)} + 3H_2O_{(l)}$.Change in gaseous moles,$\Delta n_g = n_{products} - n_{reactants} = 2 - (1 + 3.5) = -2.5$.The work done is given by $W = -\Delta n_g RT$.For $1 \ mol$ of ethane: $W = -(-2.5 	imes 8.314 	imes 300) \ J = 6235.5 \ J = 6.2355 \ kJ$.Molar mass of ethane $(C_2H_6)$ = $(2 	imes 12) + (6 	imes 1) = 30 \ g/mol = 30 	imes 10^{-3} \ kg/mol$.Given mass of ethane = $9 	imes 10^{-2} \ kg = 90 \ g$.Number of moles of ethane = $\frac{90 \ g}{30 \ g/mol} = 3 \ mol$.Total work done = $3 \ mol 	imes 6.2355 \ kJ/mol = 18.7065 \ kJ \approx 18.71 \ kJ$.

The work done during the combustion of $9 \times 10^{-2} \ kg$ of ethane,$C_2H_{6(g)}$ at $300 \ K$ is (Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,atomic mass $C = 12$,$H = 1$). (in $kJ$)

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