Calculate the standard enthalpy of reaction for the reaction 2Na + 2H2O——> 2NaOH+ H2. Standard enthalpies of formation are -285.8 kJ/mol for H2O and -470.11 kJ/mol for NaOH.
The correct answer and explanation is :
To calculate the standard enthalpy of reaction (ΔHreaction∘\Delta H^\circ_{\text{reaction}}) for the given reaction: 2Na+2H2O→2NaOH+H22 \text{Na} + 2 \text{H}_2\text{O} \rightarrow 2 \text{NaOH} + \text{H}_2
we will use the standard enthalpies of formation (ΔHf∘\Delta H^\circ_f) of the reactants and products involved. The formula to calculate the standard enthalpy of the reaction is: ΔHreaction∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H^\circ_{\text{reaction}} = \sum \Delta H^\circ_f (\text{products}) – \sum \Delta H^\circ_f (\text{reactants})
Step 1: List the standard enthalpies of formation for each compound involved.
- Standard enthalpy of formation of Na\text{Na} (sodium) is 0 kJ/mol0 \, \text{kJ/mol} (since it is in its elemental form).
- Standard enthalpy of formation of H2O\text{H}_2\text{O} (liquid water) is −285.8 kJ/mol-285.8 \, \text{kJ/mol}.
- Standard enthalpy of formation of NaOH\text{NaOH} (sodium hydroxide) is −470.11 kJ/mol-470.11 \, \text{kJ/mol}.
- Standard enthalpy of formation of H2\text{H}_2 (hydrogen gas) is 0 kJ/mol0 \, \text{kJ/mol} (since it is in its elemental form).
Step 2: Calculate the sum of the enthalpies of formation for the products and reactants.
Products:
- For 2NaOH2 \text{NaOH}, the total enthalpy is:
2×(−470.11 kJ/mol)=−940.22 kJ2 \times (-470.11 \, \text{kJ/mol}) = -940.22 \, \text{kJ}
- For H2\text{H}_2, the total enthalpy is:
1×0 kJ/mol=0 kJ1 \times 0 \, \text{kJ/mol} = 0 \, \text{kJ}
Thus, the total enthalpy for the products is: −940.22 kJ-940.22 \, \text{kJ}
Reactants:
- For 2Na2 \text{Na}, the total enthalpy is:
2×0 kJ/mol=0 kJ2 \times 0 \, \text{kJ/mol} = 0 \, \text{kJ}
- For 2H2O2 \text{H}_2\text{O}, the total enthalpy is:
2×(−285.8 kJ/mol)=−571.6 kJ2 \times (-285.8 \, \text{kJ/mol}) = -571.6 \, \text{kJ}
Thus, the total enthalpy for the reactants is: 0 kJ+(−571.6 kJ)=−571.6 kJ0 \, \text{kJ} + (-571.6 \, \text{kJ}) = -571.6 \, \text{kJ}
Step 3: Calculate the standard enthalpy of the reaction.
Now, apply the formula for the enthalpy of reaction: ΔHreaction∘=[−940.22 kJ]−[−571.6 kJ]\Delta H^\circ_{\text{reaction}} = \left[ -940.22 \, \text{kJ} \right] – \left[ -571.6 \, \text{kJ} \right] ΔHreaction∘=−940.22 kJ+571.6 kJ\Delta H^\circ_{\text{reaction}} = -940.22 \, \text{kJ} + 571.6 \, \text{kJ} ΔHreaction∘=−368.62 kJ\Delta H^\circ_{\text{reaction}} = -368.62 \, \text{kJ}
Conclusion:
The standard enthalpy of reaction for the given reaction is ΔHreaction∘=−368.62 kJ\Delta H^\circ_{\text{reaction}} = -368.62 \, \text{kJ}.
Explanation:
The standard enthalpy of reaction reflects the heat absorbed or released when the reaction occurs under standard conditions (298 K and 1 atm). In this reaction, sodium reacts with water to form sodium hydroxide and hydrogen gas. Since the enthalpy of formation for elements in their natural states (Na and H2\text{H}_2) is zero, the overall enthalpy change depends primarily on the enthalpies of formation of NaOH\text{NaOH} and H2O\text{H}_2\text{O}. The negative sign indicates that the reaction is exothermic, meaning it releases energy to the surroundings.