Question

given the data below, what is the standard enthalpy of formation of cf₄(g) in kj/mol?
ch₄(g) + 4 f₂(g) → cf₄(g) + 4 hf(g) δh°_rxn = -1645 kj

compound δh°_f (kj/mol)
ch₄(g) -75
hf(g) -260

1317 kj/mol
-680. kj/mol
-139 kj/mol
268 kj/mol

Explanation:

Step1: Recall the formula for enthalpy of reaction

$$\Delta H_{rxn}^{\circ}=\sum\Delta H_{f}^{\circ}(\text{products})-\sum\Delta H_{f}^{\circ}(\text{reactants})$$
For the reaction \(CH_{4}(g)+4F_{2}(g)\to CF_{4}(g)+4HF(g)\), \(F_{2}(g)\) is an element in its standard state, so \(\Delta H_{f}^{\circ}(F_{2}(g)) = 0\).

Step2: Substitute the known values into the formula

We know \(\Delta H_{rxn}^{\circ}=- 1645\space kJ\), \(\Delta H_{f}^{\circ}(CH_{4}(g))=-75\space kJ/mol\), \(\Delta H_{f}^{\circ}(HF(g))=-260\space kJ/mol\)

Substituting into \(\Delta H_{rxn}^{\circ}=\Delta H_{f}^{\circ}(CF_{4}(g)) + 4\Delta H_{f}^{\circ}(HF(g))-\Delta H_{f}^{\circ}(CH_{4}(g))-4\Delta H_{f}^{\circ}(F_{2}(g))\)

Since \(\Delta H_{f}^{\circ}(F_{2}(g)) = 0\), the equation becomes:

\(-1645=\Delta H_{f}^{\circ}(CF_{4}(g))+4\times(- 260)-(-75)-4\times0\)

Step3: Solve for \(\Delta H_{f}^{\circ}(CF_{4}(g))\)

First, simplify the right - hand side:

\(-1645=\Delta H_{f}^{\circ}(CF_{4}(g))-1040 + 75\)

\(-1645=\Delta H_{f}^{\circ}(CF_{4}(g))-965\)

Add \(965\) to both sides:

\(\Delta H_{f}^{\circ}(CF_{4}(g))=-1645 + 965=-680\space kJ/mol\)

Answer:

\(-680\space kJ/mol\)