Question

1. calculate the $delta_fh^{circ}$ for $so_3(g)$, given the following balanced thermochemical equation and $delta_fh^{circ}$ for $so_2(g)$ is -297 kj mol⁻¹. $2so_2(g)+o_2(g)

ightarrow2so_3(g)$ $delta_rh^{circ}=-198$ kj a) -792 kj mol⁻¹ b) -396 kj mol⁻¹ c) +198 kj mol⁻¹ d) -248 kj mol⁻¹ e) there is not enough information provided in the question.

Explanation:

Step1: Recall the formula for $\Delta_rH^{\circ}$

The formula for the enthalpy of reaction $\Delta_rH^{\circ}$ in terms of enthalpies of formation is $\Delta_rH^{\circ}=\sum n\Delta_fH^{\circ}_{products}-\sum m\Delta_fH^{\circ}_{reactants}$, where $n$ and $m$ are the stoichiometric coefficients. For the reaction $2SO_2(g)+O_2(g)
ightarrow2SO_3(g)$, $\Delta_rH^{\circ}=2\Delta_fH^{\circ}(SO_3)-2\Delta_fH^{\circ}(SO_2)-\Delta_fH^{\circ}(O_2)$. Since $\Delta_fH^{\circ}(O_2) = 0$ (the standard - enthalpy of formation of an element in its standard state is zero).

Step2: Rearrange the formula to solve for $\Delta_fH^{\circ}(SO_3)$

We know that $\Delta_rH^{\circ}=- 198\ kJ$ and $\Delta_fH^{\circ}(SO_2)=-297\ kJ/mol$. Substituting the values into the equation $\Delta_rH^{\circ}=2\Delta_fH^{\circ}(SO_3)-2\Delta_fH^{\circ}(SO_2)$ gives $-198\ kJ = 2\Delta_fH^{\circ}(SO_3)-2\times(-297\ kJ)$. First, simplify the right - hand side: $-198\ kJ=2\Delta_fH^{\circ}(SO_3)+594\ kJ$. Then, isolate $\Delta_fH^{\circ}(SO_3)$: $2\Delta_fH^{\circ}(SO_3)=-198\ kJ - 594\ kJ=-792\ kJ$. So, $\Delta_fH^{\circ}(SO_3)=\frac{-792\ kJ}{2}=-396\ kJ/mol$.

Answer:

B. $-396\ kJ\ mol^{-1}$