Question

1-Determine the amount of heat (in kJ) given off when 2.40 ×10⁴g of NO₂ is produced according to the following equation: 2 NO(g) + O₂(g) → 2NO₂(g)ΔH = −114.6 kJ/mol Enter your answer in scientific notation.

2-Given the following thermochemical equations,
NO(g) + O₃(g) → NO₂(g) + O₂(g) ΔH = −198.9 kJ/mol

O₃(g) → 3/2O₂(g) ΔH = −142.3 kJ/mol O₂(g) →2O(g) ΔH = +495 kJ/mol

determine the enthalpy change for the reaction

2NO₂(g) → 2NO(g) + O₂(g)

3-Determine the pH at 25°C of a solution prepared by dissolving 0.33 mole of ammonium chloride in 1.0 L of 0.31 M aqueous ammonia.

Answer 1

1)

Molar mass of NO2,

MM = 1*MM(N) + 2*MM(O)

= 1*14.01 + 2*16.0

= 46.01 g/mol

mass(NO2)= 2.40*10^4 g

use:

number of mol of NO2,

n = mass of NO2/molar mass of NO2

=(2.4*10^4 g)/(46.01 g/mol)

= 5.216*10^2 mol

Since Δ H is negative, heat is released

when 2 mol of NO2 reacts, heat released = 114.6 KJ

So,

for 5.216*10^2 mol of NO2, heat released = 5.216*10^2*114.6/2 KJ

= 2.989*10^4 KJ

Answer: 2.99*10^4 KJ

2)

Lets number the reaction as 1, 2, 3 from top to bottom

required reaction should be written in terms of other reaction

This is Hess Law

required reaction can be written as:

reaction 3 = -2 * (reaction 1) +2 * (reaction 2)

So, ΔHo rxn for required reaction will be:

ΔHo rxn = -2 * ΔHo rxn(reaction 1) +2 * ΔHo rxn(reaction 2)

= -2 * (-198.9) +2 * (495.0)

= 1387.8 KJ

Answer: 1387.8 KJ

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