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.

Ammonia, [tex]\( NH_3 (\Delta H = -45.9 \text{ kJ}) \)[/tex], reacts with oxygen to produce water [tex]\( (\Delta H = -241.8 \text{ kJ}) \)[/tex] and nitric oxide, [tex]\( NO (91.3 \text{ kJ}) \)[/tex], in the following reaction:

[tex]\[ 4 NH_3 (g) + 5 O_2 (g) \rightarrow 6 H_2O (g) + 4 NO (g) \][/tex]

What is the enthalpy change for this reaction?

Use [tex]\(\Delta H_{\text{reaction}} = \sum (\Delta H_{\text{products}}) - \sum (\Delta H_{\text{reactants}})\)[/tex].

A. [tex]\(-902 \text{ kJ}\)[/tex]
B. [tex]\(-104.6 \text{ kJ}\)[/tex]
C. [tex]\(104.6 \text{ kJ}\)[/tex]
D. [tex]\(900.8 \text{ kJ}\)[/tex]


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]