IDNLearn.com makes it easy to find answers and share knowledge with others. Our experts provide timely and precise responses to help you understand and solve any issue you face.
Sagot :
To determine the enthalpy change of the reaction [tex]\( H_2 (g) + Br_2 (g) \rightarrow 2 HBr (g) \)[/tex], we use the concept of standard enthalpies of formation. Here are the steps:
1. Identify the substances and their standard enthalpies of formation involved in the reaction:
From the table:
- [tex]\( \Delta H_f^\circ \)[/tex] for [tex]\( H_2 (g) = 0.0 \, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ \)[/tex] for [tex]\( Br_2 (g) = 30.907 \, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ \)[/tex] for [tex]\( HBr (g) = -36.4 \, \text{kJ/mol} \)[/tex]
2. Calculate the total enthalpy of the reactants:
The reactants are [tex]\( H_2 (g) \)[/tex] and [tex]\( Br_2 (g) \)[/tex].
- For [tex]\( H_2 (g) \)[/tex], the enthalpy is [tex]\( 0.0 \, \text{kJ/mol} \)[/tex].
- For [tex]\( Br_2 (g) \)[/tex], the enthalpy is [tex]\( 30.907 \, \text{kJ/mol} \)[/tex].
Total enthalpy of reactants:
[tex]\[ \Delta H_{\text{reactants}} = 0.0 + 30.907 = 30.907 \, \text{kJ} \][/tex]
3. Calculate the total enthalpy of the products:
The product is [tex]\( 2 \, \text{mol} \)[/tex] of [tex]\( HBr (g) \)[/tex].
- For [tex]\( HBr (g) \)[/tex], the enthalpy is [tex]\( -36.4 \, \text{kJ/mol} \)[/tex].
Total enthalpy of products:
[tex]\[ \Delta H_{\text{products}} = 2 \times (-36.4) = -72.8 \, \text{kJ} \][/tex]
4. Determine the enthalpy change of the reaction (ΔH):
[tex]\[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
Plugging in the values:
[tex]\[ \Delta H_{\text{reaction}} = -72.8 - 30.907 = -103.707 \, \text{kJ} \][/tex]
Summary:
- The total enthalpy of reactants is [tex]\( 30.907 \, \text{kJ} \)[/tex].
- The total enthalpy of products is [tex]\( -72.8 \, \text{kJ} \)[/tex].
- The enthalpy change of the reaction is [tex]\( \Delta H = -103.707 \, \text{kJ} \)[/tex].
Therefore, the enthalpy of the reaction is [tex]\( -103.707 \, \text{kJ} \)[/tex].
1. Identify the substances and their standard enthalpies of formation involved in the reaction:
From the table:
- [tex]\( \Delta H_f^\circ \)[/tex] for [tex]\( H_2 (g) = 0.0 \, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ \)[/tex] for [tex]\( Br_2 (g) = 30.907 \, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ \)[/tex] for [tex]\( HBr (g) = -36.4 \, \text{kJ/mol} \)[/tex]
2. Calculate the total enthalpy of the reactants:
The reactants are [tex]\( H_2 (g) \)[/tex] and [tex]\( Br_2 (g) \)[/tex].
- For [tex]\( H_2 (g) \)[/tex], the enthalpy is [tex]\( 0.0 \, \text{kJ/mol} \)[/tex].
- For [tex]\( Br_2 (g) \)[/tex], the enthalpy is [tex]\( 30.907 \, \text{kJ/mol} \)[/tex].
Total enthalpy of reactants:
[tex]\[ \Delta H_{\text{reactants}} = 0.0 + 30.907 = 30.907 \, \text{kJ} \][/tex]
3. Calculate the total enthalpy of the products:
The product is [tex]\( 2 \, \text{mol} \)[/tex] of [tex]\( HBr (g) \)[/tex].
- For [tex]\( HBr (g) \)[/tex], the enthalpy is [tex]\( -36.4 \, \text{kJ/mol} \)[/tex].
Total enthalpy of products:
[tex]\[ \Delta H_{\text{products}} = 2 \times (-36.4) = -72.8 \, \text{kJ} \][/tex]
4. Determine the enthalpy change of the reaction (ΔH):
[tex]\[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
Plugging in the values:
[tex]\[ \Delta H_{\text{reaction}} = -72.8 - 30.907 = -103.707 \, \text{kJ} \][/tex]
Summary:
- The total enthalpy of reactants is [tex]\( 30.907 \, \text{kJ} \)[/tex].
- The total enthalpy of products is [tex]\( -72.8 \, \text{kJ} \)[/tex].
- The enthalpy change of the reaction is [tex]\( \Delta H = -103.707 \, \text{kJ} \)[/tex].
Therefore, the enthalpy of the reaction is [tex]\( -103.707 \, \text{kJ} \)[/tex].
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.
