At room temperature, the following heats of reaction are known: eq. 1: no(g) + o3(g) → no2(g) + o2(g) δh = –198.9 kj eq. 2: 2 o3(g) → 3 o2(g) δh = –284.6 kj eq. 3: 2 o2(g) → 4 o(g) δh = –990.0 kj use the above data to calculate the heat absorbed (kj) when 2.5 moles of no2(g) is formed at room temperature according to the chemical reaction. do not include units. no(g) + o(g) → no2(g)

The heat absorbed (kJ) when 2.5 moles of NO₂ (g) is formed at room temperature is 477.25 kJ

Explanation:

Given the equations:

Equation 1:

NO(g) + O₃(g) ⇒ NO₂(g) + O₂(g) ΔH = -198.9 kJ

Equation 2:

2 O₃(g) ⇒ 3 O₂(g) ΔH = -284.6 kJ

Equation 3:

2 O₂(g) ⇒ 4 O(g) ΔH = -990.0 kJ

In order to apply Hess's law, we will reverse equation 2 and divide it by 2:

2 O₃(g) ⇒ 3 O₂(g) ΔH = -284.6 kJ

3 O₂(g) ⇒ 2 O₃(g) ΔH = +284.6 kJ

3/2 O₂(g) ⇒  O₃(g) ΔH = 284.6 kJ÷2= 142.3 kJ

Then, we reverse the equation 3 and divide it by 4:

2 O₂(g) ⇒ 4 O(g) ΔH = -990.0 kJ

4 O(g) ⇒ 2 O₂(g) ΔH = +990.0 kJ

O(g) ⇒ 1/2 O₂(g) ΔH = 990.0 kJ ÷4= 247.5 kJ

Now, we add the three equations:

NO(g) + O₃(g) ⇒ NO₂(g) + O₂(g) ΔH = -198.9 kJ

3/2 O₂(g) ⇒  O₃(g) ΔH = 142.3 kJ

O(g) ⇒ 1/2 O₂(g) ΔH = 247.5 kJ

NO(g) + O₃(g) + 3/2 O₂(g) + O(g) ⇒ NO₂(g) + O₂(g) + O₃(g) + 1/2 O₂(g)

We eliminate the terms that are repeated on both sides of the equation:

NO(g) + O(g) ⇒ NO₂(g)

The enthalpy variation is given by:

ΔH = -198.9 kJ + 142.3 kJ + 247.5 kJ = 190.9 kJ

The heat absorbed (kJ) when 2.5 moles of NO₂ (g) is formed at room temperature is:

190.9 kJ × 2.5= 477.25 kJ

your answer would be “dry”

the answer is b. dilution