Question

what is the energy change for the reaction when 1.48 g o₃ reacts with excess no?
no + o₃ → no₂ + o₂ δh = -199 kj
δe = ? kj
+2.05 kj
-6.14 kj
-4.44 kj
+1.48 kj

Explanation:

Step1: Calculate moles of \( O_3 \)

Molar mass of \( O_3 = 3\times16.00 = 48.00 \, g/mol \).
Moles of \( O_3 = \frac{mass}{molar \, mass} = \frac{1.48 \, g}{48.00 \, g/mol} \approx 0.03083 \, mol \).

Step2: Relate moles to energy change

From the reaction, 1 mol \( O_3 \) reacts with \( \Delta H = -199 \, kJ \) (exothermic, so energy change is negative).
Energy change for \( 0.03083 \, mol \) \( O_3 \):
\( \Delta E = 0.03083 \, mol \times (-199 \, kJ/mol) \approx -6.14 \, kJ \).

Answer:

-6.14 kJ (corresponding to the option: -6.14 kJ)