Ammonia $\left( NH _3(g), \Delta H_{ f }=-45.9 kJ / mol \right)$ reacts with oxygen to produce nitrogen and water $\left( H _2 O ( g ), \Delta H_{ f }=-241.8\right.
$kJ / mol$ ) according to the equation below.
$4 NH_3(g)+3 O_2(g) \rightarrow 2 N_2(g)+6 H_2 O(g)$
What is the enthalpy of the reaction?
Use $\Delta H_{\text {nan }}=\sum\left(\Delta H_{\text {f, products }}\right)-\sum\left(\Delta H_{\text {f, reectants }}\right)$.
A. -3,572.1 kJ
B. -1,267.2 kJ
C. 1,267.2 kJ
D. 3,572.1 kJ
Answer (2)
Calculate the total enthalpy of formation of the products: ∑ ( Δ H f, products ) = 2 ⋅ ( 0 kJ/mol ) + 6 ⋅ ( − 241.8 kJ/mol ) = − 1450.8 kJ .
Calculate the total enthalpy of formation of the reactants: ∑ ( Δ H f, reactants ) = 4 ⋅ ( − 45.9 kJ/mol ) + 3 ⋅ ( 0 kJ/mol ) = − 183.6 kJ .
Calculate the enthalpy of the reaction: Δ H rxn = ∑ ( Δ H f, products ) − ∑ ( Δ H f, reactants ) = − 1450.8 kJ − ( − 183.6 kJ ) = − 1267.2 kJ .
The enthalpy of the reaction is − 1267.2 kJ .
Explanation
Problem Analysis We are given the reaction: 4 N H 3 ( g ) + 3 O 2 ( g ) i g h t ha r p oo n u p 2 N 2 ( g ) + 6 H 2 O ( g ) . We are also given the enthalpies of formation for ammonia ( Δ H f ( N H 3 ( g )) = − 45.9 kJ/mol ) and water ( Δ H f ( H 2 O ( g )) = − 241.8 kJ/mol ). We need to find the enthalpy of the reaction using the formula: Δ H rxn = ∑ ( Δ H f, products ) − ∑ ( Δ H f, reactants ) .
Enthalpy of Products First, let's calculate the total enthalpy of formation for the products. The products are nitrogen ( N 2 ( g ) ) and water ( H 2 O ( g ) ). The enthalpy of formation for nitrogen is 0 kJ/mol since it is in its standard state. So, the total enthalpy of formation for the products is: ∑ ( Δ H f, products ) = 2 ⋅ Δ H f ( N 2 ( g )) + 6 ⋅ Δ H f ( H 2 O ( g )) = 2 ⋅ ( 0 kJ/mol ) + 6 ⋅ ( − 241.8 kJ/mol )
Calculating Products Enthalpy ∑ ( Δ H f, products ) = 0 + 6 ⋅ ( − 241.8 ) = − 1450.8 kJ
Enthalpy of Reactants Next, let's calculate the total enthalpy of formation for the reactants. The reactants are ammonia ( N H 3 ( g ) ) and oxygen ( O 2 ( g ) ). The enthalpy of formation for oxygen is 0 kJ/mol since it is in its standard state. So, the total enthalpy of formation for the reactants is: ∑ ( Δ H f, reactants ) = 4 ⋅ Δ H f ( N H 3 ( g )) + 3 ⋅ Δ H f ( O 2 ( g )) = 4 ⋅ ( − 45.9 kJ/mol ) + 3 ⋅ ( 0 kJ/mol )
Calculating Reactants Enthalpy ∑ ( Δ H f, reactants ) = 4 ⋅ ( − 45.9 ) + 0 = − 183.6 kJ
Enthalpy of Reaction Now, we can calculate the enthalpy of the reaction using the formula: Δ H rxn = ∑ ( Δ H f, products ) − ∑ ( Δ H f, reactants ) . Δ H rxn = − 1450.8 kJ − ( − 183.6 kJ ) = − 1450.8 + 183.6 = − 1267.2 kJ
Final Answer Therefore, the enthalpy of the reaction is − 1267.2 kJ .
Examples
The enthalpy of a reaction is crucial in various real-world applications, such as designing chemical reactors and assessing the energy efficiency of industrial processes. For instance, in the production of fertilizers, the enthalpy change of the reactions involved determines the amount of heat that needs to be supplied or removed to maintain optimal reaction conditions. Understanding these energy requirements helps engineers optimize the process, reduce energy consumption, and minimize environmental impact, leading to more sustainable and cost-effective manufacturing.
The enthalpy of the reaction of ammonia with oxygen to produce nitrogen and water is -1267.2 kJ. This value was calculated using the enthalpy of formation values for the reactants and products as per the reaction equation. Therefore, the correct choice is B. -1,267.2 kJ.
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