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7) Calculate ∆G^0' for the following reaction: glucose + 6O2 → 6 CO2 + 6 H2O. Is this reaction exergonic or endergonic?

san4es73 [151]

The reaction is endergonic.

<h2>Procedure - Determination of the Gibbs reaction type</h2>

<h3>Thermochemical model</h3>

First, we need to determine the Gibbs free energy function ( $\Delta G_{o}$ ), in kilojoules, of the entire reaction by this formula:

$\Delta G_{o} = \Sigma\limits_{j=1}^{m} n_{j}\cdot g_{o, j} - \Sigma\limits_{i=1}^{p}n_{i}\cdot g_{o,i}$ (1)

Where:

$n_{i}$ - Number of kilomoles of the i-th reactive, in kilomoles.
$n_{j}$ - Number of kilomoles of the j-th product, in kilomoles.
$g_{o,i}$ - Specific Gibbs free energy function of the i-th reactive, in kilojoules per kilomole.
$g_{o,j}$ - Specific Gibbs free energy function of the j-th reactive, in kilojoules per kilomole.

<h3>Data from thermochemical tables</h3>

From thermochemical tables we have the following information:

Reactives

$g_{glu} = -2871596\,\frac{kJ}{kmol}$ , $g_{O_{2}} = 0\,\frac{kJ}{kmol}$

$n_{glu} = 1\,kmol$ , $n_{O_{2}} = 6\,kmol$

Products

$g_{CO_{2}} = -137150\,\frac{kJ}{kmol}$ , $g_{H_{2}O} = -237180\,\frac{kJ}{kmol}$

$n_{CO_{2}} = 6\,kmol$ , $n_{H_{2}O} = 6\,kmol$

<h3>Determination of the Gibbs free energy function</h3>

Now we proceed to determine the Gibbs free energy function and later determine the nature of the <em>thermochemical</em> reaction:

$\Delta G_{o} = (6\,kmol)\cdot \left(-137150\,\frac{kJ}{kmol} \right) + (6\,kmol)\cdot \left(-237180\,\frac{kJ}{kmol} \right)-(1\,kmol)\cdot \left(-2871596\,\frac{kJ}{kmol} \right) - (6\,kmol)\cdot \left(0\,\frac{kJ}{kmol} \right)$

$\Delta G_{o} = 625616\,kJ$

The Gibbs free energy function of the <em>entire</em> reaction is 625616 kilojoules.

<h3>Analysis of the result</h3>

The Gibbs free energy function has a <em>positive</em> sign, which means that this reaction is endergonic.

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