Question

Barium chloride is sometimes used in fireworks to give a bright green color. It gives off heat when dissolved in water. $$ \mathrm{BaCl}_{2}(s) \longrightarrow \mathrm{Ba}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ (a) What is \(\Delta H\) when one mol of \(\mathrm{BaCl}_{2}\) precipitates from solution? (b) What is \(\Delta H\) when \(15.00 \mathrm{~g}\) of \(\mathrm{BaCl}_{2}\) precipitate?

Short answer

Question: Calculate the change in enthalpy for the precipitation of one mole of BaCl₂ and when 15.00 g of BaCl₂ precipitates from a solution. Answer: a) For one mole of BaCl₂ precipitation: \(\Delta H = -\Delta H_{dissolution}\) b) For 15.00 g of BaCl₂ precipitation: \(\Delta H_{15.00 g} = -\Delta H_{dissolution} \times 0.072 \, mol\)

Step-by-step solution

(Part a: Calculate \(\Delta H\) for one mol of BaCl\(_{2}\) precipitation)

To find the \(\Delta H\) for one mole of BaCl\(_{2}\) precipitation, we need to use the given reaction. As mentioned in the exercise, the dissolution of BaCl\(_{2}\) is exothermic, which means it releases heat. The precipitation of BaCl\(_{2}\) is the reverse process, so it will be endothermic, which means it will absorb heat. Let's assume the \(\Delta H\) for the dissolution process is \(\Delta H_{dissolution}\). Then, the \(\Delta H\) for the precipitation process will have the opposite sign: $$\Delta H_{precipitation} = -\Delta H_{dissolution}$$ Since we don't have a specific value for \(\Delta H_{dissolution}\), we'll denote the result as a general equation: $$\Delta H = -\Delta H_{dissolution}$$

(Part b: Calculate the \(\Delta H\) for 15.00 g BaCl\(_{2}\) precipitation)

Now we need to find the \(\Delta H\) when 15.00 g of BaCl\(_{2}\) precipitate. To do this, we first need to determine the number of moles of BaCl\(_{2}\) in 15.00 g. The molar mass of BaCl\(_{2}\) is: $$M_{BaCl_2} = 137.33 (Ba) + 2 \times 35.45 (Cl) = 137.33 + 70.90 = 208.23 \, g/mol$$ Now we can find the number of moles (n) in 15.00 g of BaCl\(_{2}\), using the formula: $$n = \frac{mass}{molar \, mass}$$ $$n = \frac{15.00 \, g}{208.23 \, g/mol} = 0.072 \, mol$$ Now we can find the \(\Delta H\) for 15.00 g of BaCl\(_{2}\) precipitation by multiplying the \(\Delta H\) per mole (our result from part a) by the number of moles for 15.00 g of BaCl\(_{2}\): $$\Delta H_{15.00 g} = \Delta H_{precipitation} \times n$$ $$\Delta H_{15.00 g} = -\Delta H_{dissolution} \times 0.072 \, mol$$ We don't have a specific value for the \(\Delta H_{dissolution}\), but with the equation above, we can calculate the \(\Delta H\) for 15.00 g of BaCl\(_{2}\) precipitation if it were provided.