Calculate the work of expansion accompanying the complete combustion of 1.0g of glucose to carbon dioxide and (a) liquid water, (b) water vapour at 20 degrees Celsiuswhen the external pressure is 1 atm
The work done for process at constant external pressure is
$$ \begin{aligned} w &=-P_{\text {ext }} \Delta V \\ &=-R T \Delta n \end{aligned} $$
Here, \(R\) is gas constant \((8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}), T\) is absolute
temperature, and \(\Delta n\) is difference between moles of gaseous
products and gaseous reactants.
(a) The combustion of glucose to carbon dioxide and liquid water
is given below:
$$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$
If \(1.0\) g glucose burns, moles of \(O_{2}\) required is \(1.0 \mathrm{~g} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \times \frac{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{180.2 \mathrm{~g}} \times \frac{6 \mathrm{~mol} \mathrm{O}_{2}}{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}=0.033 \mathrm{~mol} \mathrm{O}_{2}\)
The number of moles of \(\mathrm{CO}_{2}\) formed is
$$ \begin{array}{l} 1.0 \mathrm{~g} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \times \frac{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{180.2 \mathrm{~g}} \times \frac{6 \mathrm{~mol} \mathrm{CO}_{2}}{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}=0.033 \mathrm{~mol} \mathrm{CO}_{2} \\ \text { So, } \begin{aligned} w &=-R T\left(n_{\mathrm{CO}_{2}(g)}-n_{\mathrm{O}_{2}(g)}\right) \\ &=-R T(0.033 \mathrm{~mol}-0.033 \mathrm{~mol}) \\ &=0 \end{aligned} \end{array} $$
(b) The combustion of glucose to carbon dioxide and water vapor
is given below:
$$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$
If \(1.0 \mathrm{~g}\) glucose burns, moles of \(\mathrm{O}_{2}\) required is
ns, moles of \(\mathrm{O}_{2}\) required is
$$ 1.0 \mathrm{~g} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \times \frac{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{180.2 \mathrm{~g}} \times \frac{6 \mathrm{~mol} \mathrm{O}_{2}}{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}=0.033 \mathrm{~mol} \mathrm{O}_{2} $$
The number of moles of \(\mathrm{CO}_{2} / \mathrm{H}_{2} \mathrm{O}\) formed is
$$ \begin{array}{l} 1.0 \mathrm{~g} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \times \frac{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{180.2 \mathrm{~g}} \times \frac{6 \mathrm{~mol} \mathrm{CO}_{2}}{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}=0.033 \mathrm{~mol} \mathrm{CO}_{2} \\ 1.0 \mathrm{~g} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \times \frac{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{180.2 \mathrm{~g}} \times \frac{6 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}=0.033 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O} \\ \text { So, } w=-R T\left(n_{\infty_{2}(g)}+n_{\mathrm{H}_{2} \mathrm{O}(g)}-n_{\mathrm{O}_{2}(\mathrm{~g})}\right) \\ =-8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \times 293 \mathrm{~K}(0.033 \mathrm{~mol}+0.033 \mathrm{~mol}-0.033 \mathrm{~mol}) \\ =-80.4 \mathrm{~J} \end{array} $$
Calculate the work of expansion accompanying the complete combustion of 1.0g of glucose
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