Blake Dickson
10/14/2023 · Middle School

Calcule a energia livre de Gibbs padrão, \( \Delta, \mathrm{G}^{\circ} \), da reação: \( \mathrm{CO}_{(9)}+1 / 2 \mathrm{O}_{2(\theta)} \rightarrow \mathrm{CO}_{2(\theta)} \) à \( 25^{\circ} \mathrm{C} \) Dados: \( \Delta, \mathrm{G}^{\circ}: \mathrm{CO}_{2(g)}=-394,4 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{CO}_{(9)}=-137,2 \mathrm{~kJ} \mathrm{~mol}^{-1} \)

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Para calcular a energia livre de Gibbs padrão, \( \Delta G^{\circ} \), da reação dada, podemos usar a seguinte fórmula: \[ \Delta G^{\circ} = \sum \Delta G^{\circ}_{\text{produtos}} - \sum \Delta G^{\circ}_{\text{reagentes}} \] Onde: - \( \Delta G^{\circ} \) é a energia livre de Gibbs padrão da reação. - \( \Delta G^{\circ}_{\text{produtos}} \) é a energia livre de Gibbs padrão dos produtos. - \( \Delta G^{\circ}_{\text{reagentes}} \) é a energia livre de Gibbs padrão dos reagentes. Dadas as seguintes informações: - \( \Delta G^{\circ}_{\text{CO}_{2(g)}} = -394,4 \, \text{kJ/mol} \) - \( \Delta G^{\circ}_{\text{CO}_{(g)}} = -137,2 \, \text{kJ/mol} \) Podemos calcular \( \Delta G^{\circ} \) da reação dada da seguinte forma: \[ \Delta G^{\circ} = \Delta G^{\circ}_{\text{CO}_{2(g)}} - \Delta G^{\circ}_{\text{CO}_{(g)}} \] Vamos calcular isso. Calculate the value by following steps: - step0: Calculate: \(-394.4-\left(-137.2\right)\) - step1: Remove the parentheses: \(-394.4+137.2\) - step2: Add the numbers: \(-257.2\) A energia livre de Gibbs padrão, \( \Delta G^{\circ} \), da reação \( \text{CO}_{(g)} + \frac{1}{2} \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} \) à \( 25^{\circ} \text{C} \) é de -257,2 kJ/mol.

Quick Answer

A energia livre de Gibbs padrão, \( \Delta G^{\circ} \), da reação \( \text{CO}_{(g)} + \frac{1}{2} \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} \) à \( 25^{\circ} \text{C} \) é de -257,2 kJ/mol.
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