Calculate Δ H Rxn ∘ \Delta H_{\text{rxn}}^{\circ} Δ H Rxn ∘ ​ For Each Of The Following Reactions:(a) S I O 2 ( S ) + 4 H F ( G ) → S I F 4 ( G ) + 2 H 2 O ( L SiO_2(s) + 4HF(g) \rightarrow SiF_4(g) + 2H_2O(l S I O 2 ​ ( S ) + 4 H F ( G ) → S I F 4 ​ ( G ) + 2 H 2 ​ O ( L ](b) C 2 H 6 ( G ) + O 2 ( G ) → C O 2 ( G ) + H 2 O ( G C_2H_6(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g C 2 ​ H 6 ​ ( G ) + O 2 ​ ( G ) → C O 2 ​ ( G ) + H 2 ​ O ( G ] (unbalanced)

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Introduction

In chemistry, the enthalpy of reaction (ΔHrxn∘\Delta H_{\text{rxn}}^{\circ}) is a crucial thermodynamic property that represents the change in enthalpy that occurs during a chemical reaction. It is a measure of the energy change associated with a reaction, and it is an essential concept in understanding the spontaneity and feasibility of chemical reactions. In this article, we will discuss how to calculate ΔHrxn∘\Delta H_{\text{rxn}}^{\circ} for two given reactions.

Reaction (a): SiO2(s)+4HF(g)→SiF4(g)+2H2O(l)SiO_2(s) + 4HF(g) \rightarrow SiF_4(g) + 2H_2O(l)

To calculate ΔHrxn∘\Delta H_{\text{rxn}}^{\circ} for reaction (a), we need to know the standard enthalpies of formation (ΔHf∘\Delta H_{f}^{\circ}) of the reactants and products. The standard enthalpy of formation is the change in enthalpy that occurs when one mole of a substance is formed from its constituent elements in their standard states.

The standard enthalpies of formation for the reactants and products in reaction (a) are:

  • SiO2(s)SiO_2(s): ΔHf∘=−910.94 kJ/mol\Delta H_{f}^{\circ} = -910.94 \text{ kJ/mol}
  • HF(g)HF(g): ΔHf∘=−271.0 kJ/mol\Delta H_{f}^{\circ} = -271.0 \text{ kJ/mol}
  • SiF4(g)SiF_4(g): ΔHf∘=−1619.0 kJ/mol\Delta H_{f}^{\circ} = -1619.0 \text{ kJ/mol}
  • H2O(l)H_2O(l): ΔHf∘=−285.83 kJ/mol\Delta H_{f}^{\circ} = -285.83 \text{ kJ/mol}

Now, we can calculate ΔHrxn∘\Delta H_{\text{rxn}}^{\circ} using the following equation:

ΔHrxn∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_{\text{rxn}}^{\circ} = \sum \Delta H_{f}^{\circ}(\text{products}) - \sum \Delta H_{f}^{\circ}(\text{reactants})

Substituting the values, we get:

ΔHrxn∘=ΔHf∘(SiF4)+2ΔHf∘(H2O)−ΔHf∘(SiO2)−4ΔHf∘(HF)\Delta H_{\text{rxn}}^{\circ} = \Delta H_{f}^{\circ}(SiF_4) + 2\Delta H_{f}^{\circ}(H_2O) - \Delta H_{f}^{\circ}(SiO_2) - 4\Delta H_{f}^{\circ}(HF)

ΔHrxn∘=−1619.0 kJ/mol+2(−285.83 kJ/mol)−(−910.94 kJ/mol)−4(−271.0 kJ/mol)\Delta H_{\text{rxn}}^{\circ} = -1619.0 \text{ kJ/mol} + 2(-285.83 \text{ kJ/mol}) - (-910.94 \text{ kJ/mol}) - 4(-271.0 \text{ kJ/mol})

ΔHrxn∘=−1619.0 kJ/mol−571.66 kJ/mol+910.94 kJ/mol+1084.0 kJ/mol\Delta H_{\text{rxn}}^{\circ} = -1619.0 \text{ kJ/mol} - 571.66 \text{ kJ/mol} + 910.94 \text{ kJ/mol} + 1084.0 \text{ kJ/mol}

ΔHrxn∘=−1195.72 kJ/mol\Delta H_{\text{rxn}}^{\circ} = -1195.72 \text{ kJ/mol}

Therefore, the standard enthalpy of reaction for reaction (a) is ΔHrxn∘=−1195.72 kJ/mol\Delta H_{\text{rxn}}^{\circ} = -1195.72 \text{ kJ/mol}.

Reaction (b): C2H6(g)+O2(g)→CO2(g)+H2O(g)C_2H_6(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)

To calculate ΔHrxn∘\Delta H_{\text{rxn}}^{\circ} for reaction (b), we need to know the standard enthalpies of formation (ΔHf∘\Delta H_{f}^{\circ}) of the reactants and products. The standard enthalpy of formation is the change in enthalpy that occurs when one mole of a substance is formed from its constituent elements in their standard states.

The standard enthalpies of formation for the reactants and products in reaction (b) are:

  • C2H6(g)C_2H_6(g): ΔHf∘=−84.68 kJ/mol\Delta H_{f}^{\circ} = -84.68 \text{ kJ/mol}
  • O2(g)O_2(g): ΔHf∘=0 kJ/mol\Delta H_{f}^{\circ} = 0 \text{ kJ/mol} (by definition)
  • CO2(g)CO_2(g): ΔHf∘=−393.51 kJ/mol\Delta H_{f}^{\circ} = -393.51 \text{ kJ/mol}
  • H2O(g)H_2O(g): ΔHf∘=−241.82 kJ/mol\Delta H_{f}^{\circ} = -241.82 \text{ kJ/mol}

Now, we can calculate ΔHrxn∘\Delta H_{\text{rxn}}^{\circ} using the following equation:

ΔHrxn∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_{\text{rxn}}^{\circ} = \sum \Delta H_{f}^{\circ}(\text{products}) - \sum \Delta H_{f}^{\circ}(\text{reactants})

Substituting the values, we get:

ΔHrxn∘=ΔHf∘(CO2)+ΔHf∘(H2O)−ΔHf∘(C2H6)−ΔHf∘(O2)\Delta H_{\text{rxn}}^{\circ} = \Delta H_{f}^{\circ}(CO_2) + \Delta H_{f}^{\circ}(H_2O) - \Delta H_{f}^{\circ}(C_2H_6) - \Delta H_{f}^{\circ}(O_2)

ΔHrxn∘=−393.51 kJ/mol+(−241.82 kJ/mol)−(−84.68 kJ/mol)−0 kJ/mol\Delta H_{\text{rxn}}^{\circ} = -393.51 \text{ kJ/mol} + (-241.82 \text{ kJ/mol}) - (-84.68 \text{ kJ/mol}) - 0 \text{ kJ/mol}

ΔHrxn∘=−551.65 kJ/mol\Delta H_{\text{rxn}}^{\circ} = -551.65 \text{ kJ/mol}

Therefore, the standard enthalpy of reaction for reaction (b) is ΔHrxn∘=−551.65 kJ/mol\Delta H_{\text{rxn}}^{\circ} = -551.65 \text{ kJ/mol}.

Conclusion

In this article, we have discussed how to calculate the standard enthalpy of reaction (ΔHrxn∘\Delta H_{\text{rxn}}^{\circ}) for two given reactions. We have used the standard enthalpies of formation (ΔHf∘\Delta H_{f}^{\circ}) of the reactants and products to calculate ΔHrxn∘\Delta H_{\text{rxn}}^{\circ} using the equation:

ΔHrxn∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_{\text{rxn}}^{\circ} = \sum \Delta H_{f}^{\circ}(\text{products}) - \sum \Delta H_{f}^{\circ}(\text{reactants})

We have calculated ΔHrxn∘\Delta H_{\text{rxn}}^{\circ} for reaction (a) to be ΔHrxn∘=−1195.72 kJ/mol\Delta H_{\text{rxn}}^{\circ} = -1195.72 \text{ kJ/mol} and for reaction (b) to be ΔHrxn∘=−551.65 kJ/mol\Delta H_{\text{rxn}}^{\circ} = -551.65 \text{ kJ/mol}.

References

  • CRC Handbook of Chemistry and Physics, 97th ed. (2016)
  • Thermodynamic Data for Pure Substances, 2nd ed. (2013)
  • Standard Enthalpies of Formation, 2nd ed. (2015)

Note

Q: What is enthalpy of reaction?

A: Enthalpy of reaction (ΔHrxn∘\Delta H_{\text{rxn}}^{\circ}) is a measure of the change in enthalpy that occurs during a chemical reaction. It is a thermodynamic property that represents the energy change associated with a reaction.

Q: How do I calculate enthalpy of reaction?

A: To calculate enthalpy of reaction, you need to know the standard enthalpies of formation (ΔHf∘\Delta H_{f}^{\circ}) of the reactants and products. You can use the following equation:

ΔHrxn∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_{\text{rxn}}^{\circ} = \sum \Delta H_{f}^{\circ}(\text{products}) - \sum \Delta H_{f}^{\circ}(\text{reactants})

Q: What is the standard enthalpy of formation?

A: The standard enthalpy of formation (ΔHf∘\Delta H_{f}^{\circ}) is the change in enthalpy that occurs when one mole of a substance is formed from its constituent elements in their standard states.

Q: Where can I find the standard enthalpies of formation?

A: You can find the standard enthalpies of formation in various reference books, such as the CRC Handbook of Chemistry and Physics, or online databases, such as the National Institute of Standards and Technology (NIST) Chemistry WebBook.

Q: What are the units of enthalpy of reaction?

A: The units of enthalpy of reaction are typically kilojoules per mole (kJ/mol).

Q: Can I calculate enthalpy of reaction for a reaction that is not balanced?

A: Yes, you can calculate enthalpy of reaction for a reaction that is not balanced. However, you need to make sure that the reaction is balanced in terms of the number of moles of each substance.

Q: Can I calculate enthalpy of reaction for a reaction that involves multiple steps?

A: Yes, you can calculate enthalpy of reaction for a reaction that involves multiple steps. However, you need to make sure that the reaction is balanced in terms of the number of moles of each substance.

Q: What is the significance of enthalpy of reaction?

A: Enthalpy of reaction is an important thermodynamic property that can help you understand the spontaneity and feasibility of a chemical reaction. It can also help you predict the energy change associated with a reaction.

Q: Can I use enthalpy of reaction to predict the direction of a reaction?

A: Yes, you can use enthalpy of reaction to predict the direction of a reaction. If the enthalpy of reaction is negative, the reaction is exothermic and is likely to proceed spontaneously. If the enthalpy of reaction is positive, the reaction is endothermic and is unlikely to proceed spontaneously.

Q: Can I use enthalpy of reaction to predict the equilibrium constant of a reaction?

A: Yes, you can use enthalpy of reaction to predict the equilibrium constant of a reaction. The equilibrium constant (K) is related to the enthalpy of reaction (ΔH) by the following equation:

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, T is the temperature, and ΔS is the change in entropy.

Q: Can I use enthalpy of reaction to predict the rate of a reaction?

A: No, you cannot use enthalpy of reaction to predict the rate of a reaction. The rate of a reaction is influenced by various factors, including the concentration of reactants, the presence of catalysts, and the temperature.

Conclusion

In this article, we have answered some frequently asked questions about calculating enthalpy of reaction. We have discussed the definition of enthalpy of reaction, how to calculate it, and its significance in understanding the spontaneity and feasibility of a chemical reaction. We have also discussed some common misconceptions and limitations of using enthalpy of reaction to predict the direction and rate of a reaction.