Calculate the enthalpy of the reaction 2no(g)+o2(g)→2no2(g)
1/2n2(g)+o2(g)→no2(g), δh∘a=33.2 kj
1/2n2(g)+12o2(g)→no(g), δh∘b=90.2 kj

The enthalpy of the reaction:

2NO(g)+O₂(g)→2NO₂(g) ΔH = -114 kJ/mol

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)

Determination of the enthalpy of formation of a reaction can be through a calorimetric experiment, based on the principle of Hess's Law, enthalpy of formation table, or from bond energy data

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The value of ° H ° can be calculated from the change in enthalpy of standard formation:

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)

The elements in standard conditions are not included in the enthalpy calculations because the enthalpy of those elements under the standard conditions is zero.

From the problem can be known

∆Hf ° A: NO₂ = 33.2 kJ / mol

∆Hf ° B: NO = 90.2 kJ / mol,

In reaction:

2NO (g) + O₂ (g) → 2NO₂ (g)

Since the reaction coefficients of NO₂ and NO are 2, then ΔHf ° is also multiplied by 2

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)

∆H ° rxn = 2.∆Hf ° NO₂ - 2. ∆Hf ° NO (O₂ not included)

Delta H solution

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an exothermic reaction

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as endothermic or exothermic

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answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation:

c (diamond),

sio2 (quartz)