2. What is the enthalpy of the reaction below? $$ \mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} $$

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To determine the enthalpy of the reaction $$ \mathrm{C} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} $$, we can use the standard enthalpy of formation values for the reactants and products involved in the reaction.

1. **Identify the standard enthalpy of formation ($$ \Delta H_f^\circ $$) values:**
   - For $$ \mathrm{C} $$ (graphite): $$ \Delta H_f^\circ = 0 \, \text{kJ/mol} $$ (elements in their standard state have an enthalpy of formation of zero)
   - For $$ \mathrm{O}_{2} $$: $$ \Delta H_f^\circ = 0 \, \text{kJ/mol} $$
   - For $$ \mathrm{CO}_{2} $$: $$ \Delta H_f^\circ = -393.5 \, \text{kJ/mol} $$

2. **Use the formula for the enthalpy change of the reaction:**
   $$
   \Delta H_{reaction} = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)}
   $$

3. **Substituting the values:**
   $$
   \Delta H_{reaction} = \Delta H_f^\circ (\mathrm{CO}_{2}) - [\Delta H_f^\circ (\mathrm{C}) + \Delta H_f^\circ (\mathrm{O}_{2})]
   $$
   $$
   \Delta H_{reaction} = (-393.5 \, \text{kJ/mol}) - [0 + 0]
   $$
   $$
   \Delta H_{reaction} = -393.5 \, \text{kJ/mol}
   $$

Thus, the enthalpy of the reaction $$ \mathrm{C} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} $$ is $$ -393.5 \, \text{kJ/mol} $$. This indicates that the reaction is exothermic, releasing energy.
The enthalpy of the reaction $$ \mathrm{C} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} $$ is $$ -393.5 \, \text{kJ/mol} $$.

2. What is the enthalpy of the reaction below? \( \mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} \)

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