Question

Determine the ΔH for the following reaction 2NH3 + 5/2O2 = 2NO(g) + 3 H2O(g)

Answer 1

The enthalpy change, ΔH for the following reaction $2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)$ is -452.76 kJ.

What is enthalpy change, ΔH, of a reaction?

The enthalpy change of a reaction is the heat changes that occurs when a reaction proceeds to formation of products.

• Enthalpy change, ΔH = ΔH of products - ΔH of reactants

The equation of the reaction is given below

$2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)$

$\Delta{H_{f}\:of\:NO = 90.25 kJ; \Delta{H_{f}\:of\:H_{2}O =-241.82kJ; \Delta{H_{f}\:of\:NH_{3} =-46.1 kJ; \Delta{H_{f}\:of\:O_{2} =0$

$\Delta{H_{f}\:of\:rxn = (90.25*2)+(-241.82*3)-( -46.1*2)= -452.76\:kJ$

Therefore, the enthalpy change, ΔH for the following reaction $2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)$ is -452.76 kJ.

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Answer 2

The enthalpy change, ΔH for the following reaction

2NH₃ + 5/2O₂ --> 2NO (g) + 3 H₂O (g)  is -452.76 kJ.

What is enthalpy change, ΔH, of a reaction?

The enthalpy change of a reaction is the heat changes that occurs when a reaction proceeds to formation of products.

Formula for the enthalpy change;

ΔH = ΔH of products - ΔH of reactants

The equation of the reaction is given below

2NH₃ + 5/2O₂ --> 2NO (g) + 3 H₂O (g)

for the above reaction ;

• ΔH of NO  =  90.25 KJ
• ΔH of H₂O = - 241.82 KJ
• ΔH of NH₃ = - 46.1 KJ
• ΔH of O₂ = 0 KJ

Thus,

ΔH of Reaction = (90.25 x 2) + (-241.82 x 3) + (-46.1 x 2) = - 452.76 kJ

Therefore, the enthalpy change, ΔH for the following reaction

2NH₃ + 5/2O₂ --> 2NO (g) + 3 H₂O (g)  is - 452.76 kJ.

Learn more about enthalpy change here :

brainly.com/question/14047927

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