IDNLearn.com: Your one-stop destination for reliable answers to diverse questions. Join our community to receive timely and reliable responses to your questions from knowledgeable professionals.
Sagot :
To determine the enthalpy change for the given chemical reaction, we need to follow the enthalpy change equation:
[tex]\[ \Delta H_{\text{reaction}} = \sum (\Delta H_{\text{products}}) - \sum (\Delta H_{\text{reactants}}) \][/tex]
First, let's gather the given data for the enthalpy change of formation ([tex]\( \Delta H \)[/tex]) of each component:
- For [tex]\( NH_3 \)[/tex]: [tex]\( \Delta H_{NH_3} = -45.9 \, \text{kJ} \)[/tex]
- For [tex]\( H_2O \)[/tex]: [tex]\( \Delta H_{H_2O} = -241.8 \, \text{kJ} \)[/tex]
- For [tex]\( NO \)[/tex]: [tex]\( \Delta H_{NO} = 91.3 \, \text{kJ} \)[/tex]
Next, let's identify the coefficients from the balanced chemical reaction:
[tex]\[ 4 \, NH_3 (g) + 5 \, O_2 (g) \rightarrow 6 \, H_2O (g) + 4 \, NO (g) \][/tex]
Given the coefficients:
- 4 for [tex]\( NH_3 \)[/tex]
- 6 for [tex]\( H_2O \)[/tex]
- 4 for [tex]\( NO \)[/tex]
Now, let's calculate the total enthalpy change for the reactants:
[tex]\[ \Delta H_{\text{reactants}} = 4 \times \Delta H_{NH_3} \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = 4 \times (-45.9 \, \text{kJ}) = -183.6 \, \text{kJ} \][/tex]
Next, let's calculate the total enthalpy change for the products:
[tex]\[ \Delta H_{\text{products}} = 6 \times \Delta H_{H_2O} + 4 \times \Delta H_{NO} \][/tex]
[tex]\[ \Delta H_{\text{products}} = 6 \times (-241.8 \, \text{kJ}) + 4 \times (91.3 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -1450.8 \, \text{kJ} + 365.2 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{products}} = -1085.6 \, \text{kJ} \][/tex]
Finally, we apply the enthalpy change equation to find the enthalpy change for the reaction:
[tex]\[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -1085.6 \, \text{kJ} - (-183.6 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -1085.6 \, \text{kJ} + 183.6 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -902.0 \, \text{kJ} \][/tex]
Thus, the enthalpy change for the reaction is [tex]\( -902.0 \, \text{kJ} \)[/tex]. The correct answer is:
[tex]\[ -902 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = \sum (\Delta H_{\text{products}}) - \sum (\Delta H_{\text{reactants}}) \][/tex]
First, let's gather the given data for the enthalpy change of formation ([tex]\( \Delta H \)[/tex]) of each component:
- For [tex]\( NH_3 \)[/tex]: [tex]\( \Delta H_{NH_3} = -45.9 \, \text{kJ} \)[/tex]
- For [tex]\( H_2O \)[/tex]: [tex]\( \Delta H_{H_2O} = -241.8 \, \text{kJ} \)[/tex]
- For [tex]\( NO \)[/tex]: [tex]\( \Delta H_{NO} = 91.3 \, \text{kJ} \)[/tex]
Next, let's identify the coefficients from the balanced chemical reaction:
[tex]\[ 4 \, NH_3 (g) + 5 \, O_2 (g) \rightarrow 6 \, H_2O (g) + 4 \, NO (g) \][/tex]
Given the coefficients:
- 4 for [tex]\( NH_3 \)[/tex]
- 6 for [tex]\( H_2O \)[/tex]
- 4 for [tex]\( NO \)[/tex]
Now, let's calculate the total enthalpy change for the reactants:
[tex]\[ \Delta H_{\text{reactants}} = 4 \times \Delta H_{NH_3} \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = 4 \times (-45.9 \, \text{kJ}) = -183.6 \, \text{kJ} \][/tex]
Next, let's calculate the total enthalpy change for the products:
[tex]\[ \Delta H_{\text{products}} = 6 \times \Delta H_{H_2O} + 4 \times \Delta H_{NO} \][/tex]
[tex]\[ \Delta H_{\text{products}} = 6 \times (-241.8 \, \text{kJ}) + 4 \times (91.3 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -1450.8 \, \text{kJ} + 365.2 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{products}} = -1085.6 \, \text{kJ} \][/tex]
Finally, we apply the enthalpy change equation to find the enthalpy change for the reaction:
[tex]\[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -1085.6 \, \text{kJ} - (-183.6 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -1085.6 \, \text{kJ} + 183.6 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -902.0 \, \text{kJ} \][/tex]
Thus, the enthalpy change for the reaction is [tex]\( -902.0 \, \text{kJ} \)[/tex]. The correct answer is:
[tex]\[ -902 \, \text{kJ} \][/tex]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.