Question

3 - energy changes and rates of reactions (part 1)

question 1 (1 point)
the combustion of methane releases 890 kj/mol. this means:
the reaction is endothermic
δh = +890 kj/mol
δh = -890 kj/mol
energy is absorbed by methane
question 2 (1 point)
given the thermochemical equation n₂(g) + 2o₂(g) + 66.4 kj --> 2 no₂(g), what would you expect the molar change of enthalpy (δhᵣ) of the decomposition of no₂(g) into nitrogen gas and oxygen gas to be?
33.2 kj
-66.4 kj
-33.2 kj
66.4 kj

Explanation:

Question 1

• Combustion reactions release energy, so they are exothermic. In exothermic reactions, the enthalpy change ($\Delta H$) is negative because the system (reactants) loses energy to the surroundings.
• "The reaction is endothermic" is wrong as endothermic reactions absorb energy.
• "$\Delta H = +890$ kJ/mol" is for endothermic (energy absorbed), not exothermic (energy released).
• "Energy is absorbed by methane" is incorrect because combustion releases energy from methane.
• So the correct statement is $\Delta H = -890$ kJ/mol as the reaction releases energy (exothermic, negative $\Delta H$).

1. First, analyze the given reaction: $\ce{N_{2(g)} + 2O_{2(g)} + 66.4\ kJ -> 2NO_{2(g)}}$. This reaction is endothermic (energy is a reactant, so $\Delta H$ for this reaction is $+66.4$ kJ/mol as energy is absorbed to form $\ce{NO_{2(g)}}$).
2. The decomposition of $\ce{NO_{2(g)}}$ is the reverse reaction: $\ce{2NO_{2(g)} -> N_{2(g)} + 2O_{2(g)}}$. For a reverse reaction, the enthalpy change is the negative of the forward reaction's enthalpy change. Also, if we want the molar enthalpy change for the decomposition of 1 mole of $\ce{NO_{2(g)}}$, we divide the enthalpy change of the reverse reaction by 2.
3. The forward reaction (formation of $\ce{2NO_{2(g)}}$) has $\Delta H = +66.4$ kJ/mol. So the reverse reaction (decomposition of $\ce{2NO_{2(g)}}$) has $\Delta H = -66.4$ kJ/mol. For the decomposition of 1 mole of $\ce{NO_{2(g)}}$, we calculate $\frac{-66.4\ \text{kJ/mol}}{2} = -33.2\ \text{kJ/mol}$.

Answer:

$\boldsymbol{\Delta H = -890}$ kJ/mol (the option with $\Delta H = -890$ kJ/mol)

Question 2