Ammonia, [tex]$NH _3(\Delta H_1=-45.9 kJ)$[/tex], reacts with oxygen to produce water [tex]$\left(\Delta H_1=-241.8 kJ\right)$[/tex] and nitric oxide, [tex]$NO \left(\Delta H_1=\right.$ 91.3 kJ )[/tex], in the following reaction:
[tex]$4 NH_3(g)+5 O_2(g) \rightarrow 6 H_2 O(g)+4 NO(g)$[/tex]
What is the enthalpy change for this reaction?
Use [tex]$\Delta H_{\text {rxn }}=\sum\left(\Delta H_{\text {t, products }}\right)-\sum\left(\Delta H_{\text {f, reactants }}\right)$[/tex].
A. -902 kJ
B. -104.6 kJ
C. 104.6 kJ
D. 900.8 kJ
Answer (2)
Calculate the total enthalpy of formation for the products: ∑ Δ H f , p ro d u c t s = 6 × ( − 241.8 ) + 4 × ( 91.3 ) = − 1085.6 kJ/mol .
Calculate the total enthalpy of formation for the reactants: ∑ Δ H f , re a c t an t s = 4 × ( − 45.9 ) + 5 × ( 0 ) = − 183.6 kJ/mol .
Calculate the enthalpy change for the reaction: Δ H r x n = − 1085.6 − ( − 183.6 ) = − 902.0 kJ/mol .
The enthalpy change for the reaction is − 902 kJ .
Explanation
Understanding the Problem We are given the reaction: 4 N H 3 ( g ) + 5 O 2 ( g ) i g h t ha r p oo n u p 6 H 2 O ( g ) + 4 NO ( g ) . We are also given the enthalpies of formation for ammonia ( N H 3 ), water ( H 2 O ), and nitric oxide ( NO ). We need to find the enthalpy change for the reaction using the formula: Δ H r x n = ∑ ( Δ H f , p ro d u c t s ) − ∑ ( Δ H f , re a c t an t s ) .
Calculating Enthalpy of Products First, we calculate the total enthalpy of formation for the products. This includes 6 moles of water and 4 moles of nitric oxide: ∑ Δ H f , p ro d u c t s = 6 × Δ H f ( H 2 O ) + 4 × Δ H f ( NO ) ∑ Δ H f , p ro d u c t s = 6 × ( − 241.8 kJ/mol ) + 4 × ( 91.3 kJ/mol ) ∑ Δ H f , p ro d u c t s = − 1450.8 kJ/mol + 365.2 kJ/mol ∑ Δ H f , p ro d u c t s = − 1085.6 kJ/mol
Calculating Enthalpy of Reactants Next, we calculate the total enthalpy of formation for the reactants. This includes 4 moles of ammonia and 5 moles of oxygen: ∑ Δ H f , re a c t an t s = 4 × Δ H f ( N H 3 ) + 5 × Δ H f ( O 2 ) ∑ Δ H f , re a c t an t s = 4 × ( − 45.9 kJ/mol ) + 5 × ( 0 kJ/mol ) ∑ Δ H f , re a c t an t s = − 183.6 kJ/mol + 0 kJ/mol ∑ Δ H f , re a c t an t s = − 183.6 kJ/mol
Calculating Enthalpy Change of Reaction Now, we can calculate the enthalpy change for the reaction: Δ H r x n = ∑ Δ H f , p ro d u c t s − ∑ Δ H f , re a c t an t s Δ H r x n = − 1085.6 kJ/mol − ( − 183.6 kJ/mol ) Δ H r x n = − 1085.6 kJ/mol + 183.6 kJ/mol Δ H r x n = − 902.0 kJ/mol
Final Answer The enthalpy change for the reaction is -902.0 kJ.
Examples
Understanding enthalpy changes is crucial in many real-world applications, such as designing efficient combustion engines or optimizing industrial chemical processes. For example, when designing a car engine, engineers need to know the enthalpy change of the fuel combustion reaction to determine how much energy will be released. This helps them optimize the engine's design for maximum efficiency and minimal environmental impact. Similarly, in the production of fertilizers, understanding the enthalpy changes of various reactions allows for the development of energy-efficient and cost-effective manufacturing processes.
The enthalpy change for the reaction of ammonia with oxygen to produce water and nitric oxide is calculated to be -902 kJ. This value indicates that the reaction is exothermic, releasing energy. Therefore, the correct answer is A . -902 kJ.
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