Answer :
To determine the enthalpy change of a reaction ([tex]\( \Delta H_{\text{rxn}} \)[/tex]), we need the balanced chemical equation for the reaction and the standard enthalpies of formation ([tex]\( \Delta H_f^\circ \)[/tex]) for the reactants and products. Unfortunately, your question does not provide the balanced chemical equation. However, I can explain the general steps to calculate the enthalpy change of a reaction using the standard enthalpies of formation.
Step-by-Step Solution:
1. Write the Balanced Chemical Equation:
- The first step is to write the balanced equation for the reaction. For example, consider a hypothetical reaction:
[tex]\[ aA + bB \rightarrow cC + dD \][/tex]
2. Identify the Standard Enthalpies of Formation:
- From the provided table, identify the [tex]\( \Delta H_f^\circ \)[/tex] values for each reactant (A, B) and product (C, D).
3. Calculate the Enthalpy Change for Reactants and Products:
- Multiply the standard enthalpy of formation of each reactant and product by their respective coefficients from the balanced chemical equation.
[tex]\[ \Delta H_{\text{reactants}} = a \Delta H_f^\circ(A) + b \Delta H_f^\circ(B) \][/tex]
[tex]\[ \Delta H_{\text{products}} = c \Delta H_f^\circ(C) + d \Delta H_f^\circ(D) \][/tex]
4. Apply Hess's Law to Calculate the Reaction Enthalpy:
- Use the relation:
[tex]\[ \Delta H_{\text{rxn}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
Example Calculation:
Given the table values:
- [tex]\( \Delta H_f^\circ(C_2H_2(g)) = -26.7 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ(NH_3(g)) = -46.19 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ(HBr(g)) = 236.23 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ(HCl(g)) = -92.30 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ(HF(g)) = -268.6 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ(HI(g)) = 25.9 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ(NaCl(s)) = -411.0 \text{ kJ/mol} \)[/tex]
Let's assume we have a balanced chemical equation (hypothetical example):
[tex]\[ 2 HCl(g) + Na_2O(s) \rightarrow 2 NaCl(s) + H_2O(g) \][/tex]
- Identify [tex]\( \Delta H_f^\circ \)[/tex] for each component from the table:
- [tex]\( HCl(g) \)[/tex]: [tex]\( -92.30 \text{ kJ/mol} \)[/tex]
- [tex]\( NaCl(s) \)[/tex]: [tex]\( -411.0 \text{ kJ/mol} \)[/tex]
- As [tex]\( Na_2O(s) \)[/tex] and [tex]\( H_2O(g) \)[/tex] are not provided, we'll use hypothetical values:
- [tex]\( Na_2O(s) \)[/tex]: [tex]\( -256.7 \text{ kJ/mol} \)[/tex]
- [tex]\( H_2O(g) \)[/tex]: [tex]\( -241.8 \text{ kJ/mol} \)[/tex]
- Calculate the total enthalpy for reactants:
[tex]\[ \Delta H_{\text{reactants}} = 2 \cdot (-92.30) + (-256.7) = -184.6 - 256.7 = -441.3 \text{ kJ/mol} \][/tex]
- Calculate the total enthalpy for products:
[tex]\[ \Delta H_{\text{products}} = 2 \cdot (-411.0) + (-241.8) = -822.0 - 241.8 = -1063.8 \text{ kJ/mol} \][/tex]
- Apply Hess's Law:
[tex]\[ \Delta H_{\text{rxn}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} = -1063.8 \text{ kJ/mol} - (-441.3 \text{ kJ/mol}) = -1063.8 + 441.3 = -622.5 \text{ kJ/mol} \][/tex]
This is an illustrative example; the actual calculation would depend entirely on the correct balanced chemical equation and the provided data.
Step-by-Step Solution:
1. Write the Balanced Chemical Equation:
- The first step is to write the balanced equation for the reaction. For example, consider a hypothetical reaction:
[tex]\[ aA + bB \rightarrow cC + dD \][/tex]
2. Identify the Standard Enthalpies of Formation:
- From the provided table, identify the [tex]\( \Delta H_f^\circ \)[/tex] values for each reactant (A, B) and product (C, D).
3. Calculate the Enthalpy Change for Reactants and Products:
- Multiply the standard enthalpy of formation of each reactant and product by their respective coefficients from the balanced chemical equation.
[tex]\[ \Delta H_{\text{reactants}} = a \Delta H_f^\circ(A) + b \Delta H_f^\circ(B) \][/tex]
[tex]\[ \Delta H_{\text{products}} = c \Delta H_f^\circ(C) + d \Delta H_f^\circ(D) \][/tex]
4. Apply Hess's Law to Calculate the Reaction Enthalpy:
- Use the relation:
[tex]\[ \Delta H_{\text{rxn}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
Example Calculation:
Given the table values:
- [tex]\( \Delta H_f^\circ(C_2H_2(g)) = -26.7 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ(NH_3(g)) = -46.19 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ(HBr(g)) = 236.23 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ(HCl(g)) = -92.30 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ(HF(g)) = -268.6 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ(HI(g)) = 25.9 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ(NaCl(s)) = -411.0 \text{ kJ/mol} \)[/tex]
Let's assume we have a balanced chemical equation (hypothetical example):
[tex]\[ 2 HCl(g) + Na_2O(s) \rightarrow 2 NaCl(s) + H_2O(g) \][/tex]
- Identify [tex]\( \Delta H_f^\circ \)[/tex] for each component from the table:
- [tex]\( HCl(g) \)[/tex]: [tex]\( -92.30 \text{ kJ/mol} \)[/tex]
- [tex]\( NaCl(s) \)[/tex]: [tex]\( -411.0 \text{ kJ/mol} \)[/tex]
- As [tex]\( Na_2O(s) \)[/tex] and [tex]\( H_2O(g) \)[/tex] are not provided, we'll use hypothetical values:
- [tex]\( Na_2O(s) \)[/tex]: [tex]\( -256.7 \text{ kJ/mol} \)[/tex]
- [tex]\( H_2O(g) \)[/tex]: [tex]\( -241.8 \text{ kJ/mol} \)[/tex]
- Calculate the total enthalpy for reactants:
[tex]\[ \Delta H_{\text{reactants}} = 2 \cdot (-92.30) + (-256.7) = -184.6 - 256.7 = -441.3 \text{ kJ/mol} \][/tex]
- Calculate the total enthalpy for products:
[tex]\[ \Delta H_{\text{products}} = 2 \cdot (-411.0) + (-241.8) = -822.0 - 241.8 = -1063.8 \text{ kJ/mol} \][/tex]
- Apply Hess's Law:
[tex]\[ \Delta H_{\text{rxn}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} = -1063.8 \text{ kJ/mol} - (-441.3 \text{ kJ/mol}) = -1063.8 + 441.3 = -622.5 \text{ kJ/mol} \][/tex]
This is an illustrative example; the actual calculation would depend entirely on the correct balanced chemical equation and the provided data.
