Question

Ammonia, NH3 (Delta. Hf = –46. 2 kJ), reacts with oxygen to produce water (Delta. Hf = –241. 8 kJ) and nitric oxide, NO (Delta. Hf = 91. 3 kJ), in the following reaction: 4 upper N upper H subscript 3 (g) plus 5 upper O subscript 2 (g) right arrow 6 upper H subscript 2 upper O (g) plus 4 upper N upper O (g). What is the enthalpy change for this reaction? Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants. -900. 8 kJ –104. 6 kJ 104. 6 kJ 900. 8 kJ.

Answer 1

The change in enthalpy for the reaction has been 900.8 kJ. Thus, option D is correct.

Enthalpy change has been defined as the amount of heat energy produced, or absorbed or released by the reactants in the chemical reaction for the formation of products.

Computation for the Enthalpy change of reaction

The given chemical reaction has been:

$\rm 4\;NH_3\;(g)\;+\;5\;O_2\;(g)\;\rightarrow\;6\;H_2O\;(g)\;+\;4\;NO\;(g)$

The change in enthalpy in the reaction has been given as:

$\Delta H_{rxn}=\Delta H_{Product}-\Delta H_{Reactant}$

The enthalpy for products and reactants has been given in the table attached.

The change in enthalpy for reaction has been given by substituting the values as:

$\Delta H_{rxn}=6\;\times\;\Delta H (\rm{H_2O} )\;+\;4\;\times\;\Delta H (\text{NO})-4\;\times\;\Delta H(\rm{NH_3})\\\Delta \textit H_{\textit {rxn}}=6\;\times\;(-241.8)\;+\;4\;\times\;(91.3)\;-\;4\;\times\;(-46.2)\;kJ\\\Delta \textit H_{\textit {rxn}}=900.8\;kJ$

The change in enthalpy for the reaction has been 900.8 kJ. Thus, option D is correct.

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