Ammonia, [tex]$NH _3(\Delta H_{ f }=-45.9 kJ)$[/tex], reacts with oxygen to produce water ([tex]$\Delta H_{ f }=-241.8 kJ$[/tex]) and nitric oxide, [tex]$NO (\Delta H_{ f }=$[/tex] 91.3 kJ ), 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]$\Delta H_{r x n}=\sum(\Delta H_{f, \text { products }})-\sum(\Delta H_{f, \text { reactants }})$[/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: 6 × ( − 241.8 k J ) + 4 × ( 91.3 k J ) = − 1085.6 k J .
Calculate the total enthalpy of formation for the reactants: 4 × ( − 45.9 k J ) + 5 × ( 0 k J ) = − 183.6 k J .
Calculate the enthalpy change of the reaction: − 1085.6 k J − ( − 183.6 k J ) = − 902 k J .
The enthalpy change for the reaction is − 902 k J .
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 each compound: Δ H f ( N H 3 ) = − 45.9 k J Δ H f ( H 2 O ) = − 241.8 k J Δ H f ( NO ) = 91.3 k J Δ H f ( O 2 ) = 0 k J (since it is an element in its standard state)
The formula to calculate the enthalpy change of the reaction is: \Delta H_{rxn} = \sum \Delta H_{f, products} - \sum \Delta H_{f, reactants}$ 2. Calculating Enthalpy of Products First, calculate the total enthalpy of formation of the products: \sum \Delta H_{f, products} = 6 \cdot \Delta H_{f}(H_2O) + 4 \cdot \Delta H_{f}(NO) \sum \Delta H_{f, products} = 6 \cdot (-241.8 kJ) + 4 \cdot (91.3 kJ) \sum \Delta H_{f, products} = -1450.8 kJ + 365.2 kJ = -1085.6 kJ 3. C a l c u l a t in g E n t ha lp yo f R e a c t an t s N e x t , c a l c u l a t e t h e t o t a l e n t ha lp yo ff or ma t i o n o f t h ere a c t an t s : \sum \Delta H_{f, reactants} = 4 \cdot \Delta H_{f}(NH_3) + 5 \cdot \Delta H_{f}(O_2) \sum \Delta H_{f, reactants} = 4 \cdot (-45.9 kJ) + 5 \cdot (0 kJ) \sum \Delta H_{f, reactants} = -183.6 kJ + 0 kJ = -183.6 kJ 4. C a l c u l a t in g E n t ha lp y C han g eo f R e a c t i o n N o w , c a l c u l a t e t h ee n t ha lp yc han g eo f t h ere a c t i o n u s in g t h e f or m u l a : \Delta H_{rxn} = \sum \Delta H_{f, products} - \sum \Delta H_{f, reactants} \Delta H_{rxn} = -1085.6 kJ - (-183.6 kJ) \Delta H_{rxn} = -1085.6 kJ + 183.6 kJ = -902 kJ
Final Answer Therefore, the enthalpy change for this reaction is − 902 k J .
Examples
This calculation is crucial in industrial processes, especially in the production of nitric acid, a key component in fertilizers. By understanding the enthalpy change, engineers can optimize reaction conditions to maximize efficiency and minimize energy consumption, leading to cost savings and reduced environmental impact. This principle extends to various chemical reactions, aiding in designing energy-efficient and sustainable processes.
The enthalpy change for the reaction of ammonia with oxygen to produce water and nitric oxide is calculated to be -902 kJ. This is determined using the enthalpy of formation values for the products and reactants. Therefore, the correct answer is A. -902 kJ.
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