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3 years ago

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The combustion of 5.00 grams of c2h6(g), at constant pressure releases 259 kj of heat. what is δh for the reaction: 2c2h6(g) + 7

o2(g) => 4co2(g) + 6h2o(l)?

1 answer:

Vlad1618 [11]3 years ago

7 0

The molar mass of C2H6 is 28 g/mol, so the number of moles is:

moles C2H6 = 5 / 28 = 0.179 mol

From the equation, there are 2 moles of C2H6, therefore the heat of reaction is:

δh = (259 kj / 0.179 mol) * 2 mol

<span>δh = 2900 kJ = 2.9 MJ</span>

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a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

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$x_{Ne}=\frac{x mol}{12 mol}=x/12$

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Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

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$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

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