Combustion of 1 mole of benzene is expressed at

$$\mathrm{C}_6 \mathrm{H}_6(\mathrm{l})+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \text {. }$$

The standard enthalpy of combustion of $2 \mathrm{~mol}$ of benzene is $-^{\prime} x^{\prime} \mathrm{kJ}$. $x=$ __________.

Given :

1. standard Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{C}_6 \mathrm{H}_6(\mathrm{l})$, for the reaction $6 \mathrm{C}$ (graphite) $+3 \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{C}_6 \mathrm{H}_6(\mathrm{l})$ is $48.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

2. Standard Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{CO}_2(\mathrm{g})$, for the reaction $\mathrm{C}$ (graphite) $+\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g})$ is $-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

3. Standard and Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$, for the reaction $$\mathrm{H}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})$$ is $-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

<p>To determine the standard enthalpy of combustion of 2 moles of benzene, we need to use the standard enthalpy of formation values provided and apply Hess's Law. Here is a step-by-step explanation:</p> <h3>Given Data:</h3> <ol> <li><strong>Standard enthalpy of formation of benzene ($C_6H_6(l)$):</strong></li> </ol> <p>$ \Delta H_f(\text{C}_6\text{H}_6(l)) = 48.5 \, \text{kJ/mol} $</p> <ol> <li><strong>Standard enthalpy of formation of carbon dioxide ($CO_2(g)$):</strong></li> </ol> <p>$ \Delta H_f(\text{CO}_2(g)) = -393.5 \, \text{kJ/mol} $</p> <ol> <li><strong>Standard enthalpy of formation of water ($H_2O(l)$):</strong></li> </ol> <p>$ \Delta H_f(\text{H}_2O(l)) = -286 \, \text{kJ/mol} $</p> <h3>Combustion Reaction for Benzene:</h3> <p>$ \text{C}_6\text{H}_6(l) + \frac{15}{2} \text{O}_2(g) \rightarrow 6 \text{CO}_2(g) + 3 \text{H}_2O(l) $</p> <h3>Enthalpy Change Calculation:</h3> <p>Using Hess's Law, the enthalpy change for the reaction can be calculated as follows:</p> <p>$ \Delta H_{\text{comb}} = \left[ 6 \Delta H_f(\text{CO}_2(g)) + 3 \Delta H_f(\text{H}_2O(l)) \right] - \Delta H_f(\text{C}_6\text{H}_6(l)) $</p> <p>Substitute the given values:</p> <p>$ \Delta H_{\text{comb}} = \left[ 6 \times (-393.5) + 3 \times (-286) \right] - 48.5 $</p> <p>Perform the calculations:</p> <p>$ \Delta H_{\text{comb}} = \left[ 6 \times (-393.5) \right] + \left[ 3 \times (-286) \right] - 48.5 $</p> <p>$ \Delta H_{\text{comb}} = \left[ -2361 \right] + \left[ -858 \right] - 48.5 $</p> <p>$ \Delta H_{\text{comb}} = -3267.5 \, \text{kJ/mol} $</p> <p>This value is the enthalpy change for the combustion of 1 mole of benzene.</p> <h3>For 2 Moles of Benzene:</h3> <p>$ \Delta H_{\text{comb (2 moles)}} = 2 \times (-3267.5 \, \text{kJ/mol}) $</p> <p>$ \Delta H_{\text{comb (2 moles)}} = -6535 \, \text{kJ} $</p> <h3>Conclusion:</h3> <p>The standard enthalpy of combustion of 2 moles of benzene is</p> <p>$ x = 6535 \, \text{kJ} $</p> <p>Thus, $ x = 6535 $.</p>