copper (II) sulfate reacts with sodium hydroxide in a double displacement reaction

copper (II) sulfate reacts with sodium hydroxide in a double displacement reaction. 50 mL of 0.30 M CuSO4 (aq)Ã‚Â is mixed with and equal volume of 0.699 M of NaOH(aq) solution. the initial temperature of bith solutions was 21.4 degree. after mixing the solutions in a styroform calorimeter, the highest temperature that is reached is 24.6 degree. determine the molar enthalpy for copper (II) sulfate. write the balanced chemical equation and include the energy as a term in the equation. we can the heat capacity of water as the both solutions are aqueous. which is 4.19 J\g.C

The Correct Answer and Explanation is :

To determine the molar enthalpy ((\Delta H)) for copper(II) sulfate ((\ce{CuSO4})), follow these steps:

Step 1: Write the Balanced Chemical Equation

The reaction between (\ce{CuSO4}) and (\ce{NaOH}) is a double displacement reaction, forming copper(II) hydroxide precipitate and sodium sulfate:

[
\ce{CuSO4 (aq) + 2NaOH (aq) -> Cu(OH)2 (s) + Na2SO4 (aq)}
]

Since this reaction releases energy, it is exothermic.

Step 2: Calculate the Heat Energy Released

The heat ((q)) released in the reaction is calculated using the formula:

[
q = mc\Delta T
]

where:

(m) is the mass of the solution (assuming the density of the solution is similar to water, 1.00 g/mL),
(c) is the specific heat capacity of water ((4.19 \, \text{J/g°C})),
(\Delta T) is the temperature change.

Given:

Initial temperature: 21.4°C
Final temperature: 24.6°C
Volume of (\ce{CuSO4}) solution: 50 mL
Volume of (\ce{NaOH}) solution: 50 mL
Total volume of solution: 100 mL
(\Delta T = 24.6 – 21.4 = 3.2°C)
Since 100 mL of solution is approximately 100 g (assuming water density),

[
q = (100 \, \text{g}) (4.19 \, \text{J/g°C}) (3.2 \, \text{°C})
]

[
q = 1340.8 \, \text{J} \approx 1.34 \, \text{kJ}
]

Step 3: Determine the Limiting Reactant

Moles of (\ce{CuSO4}):

[
n_{\ce{CuSO4}} = 0.30 \, M \times 0.050 \, L = 0.015 \, \text{mol}
]

Moles of (\ce{NaOH}):

[
n_{\ce{NaOH}} = 0.699 \, M \times 0.050 \, L = 0.03495 \, \text{mol}
]

Since (\ce{CuSO4}) reacts with (\ce{NaOH}) in a 1:2 ratio, the amount of (\ce{NaOH}) needed for 0.015 mol (\ce{CuSO4}) is:

[
0.015 \times 2 = 0.030 \text{ mol NaOH}
]

Since there is 0.03495 mol of (\ce{NaOH}) available, which is more than required, (\ce{CuSO4}) is the limiting reactant.

Step 4: Calculate the Molar Enthalpy Change

[
\Delta H = \frac{q}{n_{\ce{CuSO4}}}
]

[
\Delta H = \frac{1.34 \, \text{kJ}}{0.015 \, \text{mol}}
]

[
\Delta H = -89.33 \, \text{kJ/mol}
]

(Since heat is released, (\Delta H) is negative, indicating an exothermic reaction.)

Step 5: Write the Balanced Equation with Energy

[
\ce{CuSO4 (aq) + 2NaOH (aq) -> Cu(OH)2 (s) + Na2SO4 (aq) \quad \Delta H = -89.3 \, kJ/mol}
]

Explanation (300 Words)

This reaction is a double displacement reaction where copper(II) sulfate ((\ce{CuSO4})) reacts with sodium hydroxide ((\ce{NaOH})) to form copper(II) hydroxide ((\ce{Cu(OH)2})), a solid precipitate, and sodium sulfate ((\ce{Na2SO4})), which remains in solution. Since heat is released in this reaction, it is an exothermic reaction.

To determine the molar enthalpy change, we first calculate the heat energy released using the formula (q = mc\Delta T). Assuming that the solution behaves like water, we take the total mass as 100 g (since 1 mL ≈ 1 g), the specific heat capacity as 4.19 J/g°C, and the temperature change as (3.2^\circ C). This gives us a heat release of 1.34 kJ.

Next, we find the limiting reactant. (\ce{CuSO4}) is the limiting reactant since it fully reacts with excess (\ce{NaOH}). We then use the moles of (\ce{CuSO4}) (0.015 mol) to calculate the enthalpy change per mole:

[
\Delta H = \frac{1.34 \, \text{kJ}}{0.015 \, \text{mol}} = -89.3 \, \text{kJ/mol}
]

This means that when one mole of (\ce{CuSO4}) reacts with sodium hydroxide, 89.3 kJ of energy is released. The negative sign indicates that the reaction is exothermic.

Thus, the balanced reaction including energy is: