Question

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

Answer 1

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

$$ \begin{array}{l} 1.0 \mathrm{~g} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \times \frac{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{180.2 \mathrm{~g}} \times \frac{6 \mathrm{~mol} \mathrm{CO}_{2}}{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}=0.033 \mathrm{~mol} \mathrm{CO}_{2} \\ 1.0 \mathrm{~g} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \times \frac{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{180.2 \mathrm{~g}} \times \frac{6 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}=0.033 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O} \\ \text { So, } w=-R T\left(n_{\infty_{2}(g)}+n_{\mathrm{H}_{2} \mathrm{O}(g)}-n_{\mathrm{O}_{2}(\mathrm{~g})}\right) \\ =-8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \times 293 \mathrm{~K}(0.033 \mathrm{~mol}+0.033 \mathrm{~mol}-0.033 \mathrm{~mol}) \\ =-80.4 \mathrm{~J} \end{array} $$