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 $\Delta H_{\text {rxn }}=\sum\left(\Delta H_{\text {f,products }}\right)-\sum\left(\Delta H_{f, \text { reactants }}\right)$.
A. -3,572.1 kJ
B. -1,267.2 kJ
C. 1,267.2 kJ
D. 3,572.1 kJ
Answer (1)
Calculate the total enthalpy of formation for the products: ∑ Δ H f ( p ro d u c t s ) = 2 × ( 0 m o l k J ) + 6 × ( − 241.8 m o l k J ) = − 1450.8 m o l k J .
Calculate the total enthalpy of formation for the reactants: ∑ Δ H f ( re a c t an t s ) = 4 × ( − 45.9 m o l k J ) + 3 × ( 0 m o l k J ) = − 183.6 m o l k J .
Calculate the enthalpy of the reaction: Δ H r x n = − 1450.8 m o l k J − ( − 183.6 m o l k J ) = − 1267.2 m o l k J .
The enthalpy of the reaction is − 1267.2 k J .
Explanation
Problem Analysis and Given Data We are given the balanced chemical equation: 4 N H 3 ( g ) + 3 O 2 ( g ) → 2 N 2 ( g ) + 6 H 2 O ( g ) We are also given the enthalpy of formation for ammonia ( N H 3 ) as − 45.9 k J / m o l and for water ( H 2 O ) as − 241.8 k J / m o l . We need to find the enthalpy of 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 Formation for Products First, let's calculate the total enthalpy of formation for the products. The products are 2 N 2 ( g ) and 6 H 2 O ( g ) . The enthalpy of formation for N 2 ( g ) is 0 k J / m o l because it is in its standard state. ∑ Δ H f ( p ro d u c t s ) = 2 × Δ H f ( N 2 ( g )) + 6 × Δ H f ( H 2 O ( g )) ∑ Δ H f ( p ro d u c t s ) = 2 × ( 0 m o l k J ) + 6 × ( − 241.8 m o l k J ) ∑ Δ H f ( p ro d u c t s ) = 0 m o l k J − 1450.8 m o l k J = − 1450.8 m o l k J
Calculating Enthalpy of Formation for Reactants Next, let's calculate the total enthalpy of formation for the reactants. The reactants are 4 N H 3 ( g ) and 3 O 2 ( g ) . The enthalpy of formation for O 2 ( g ) is 0 k J / m o l because it is in its standard state. ∑ Δ H f ( re a c t an t s ) = 4 × Δ H f ( N H 3 ( g )) + 3 × Δ H f ( O 2 ( g )) ∑ Δ H f ( re a c t an t s ) = 4 × ( − 45.9 m o l k J ) + 3 × ( 0 m o l k J ) ∑ Δ H f ( re a c t an t s ) = − 183.6 m o l k J + 0 m o l k J = − 183.6 m o l k J
Calculating Enthalpy of Reaction Now, we can calculate the enthalpy of 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 = − 1450.8 m o l k J − ( − 183.6 m o l k J ) Δ H r x n = − 1450.8 m o l k J + 183.6 m o l k J = − 1267.2 m o l k J
Final Answer The enthalpy of the reaction is − 1267.2 k J / m o l .
Examples
The concept of enthalpy of reaction is crucial in various real-world applications, especially in chemical engineering and industrial processes. For instance, when designing a chemical plant that produces ammonia-based fertilizers, engineers need to accurately calculate the enthalpy change to manage the heat released or absorbed during the reaction. This ensures the process is energy-efficient and safe, preventing overheating or cooling issues that could affect the yield and quality of the fertilizer. Understanding enthalpy changes helps optimize reaction conditions, making the production process economically viable and environmentally sustainable.
