Question

What is the energy change when 249 mL of 15.0 M nitric acid is diluted with D.I. water? If this nitric acid is added to 322 mL of D.I. water at an initial temperature of 21.4 °C, what will be the final temperature of the resulting solution? HNO₃ (15.0 M) HNO → HNO₃(dilute) ΔH°= -6.88 kJ/mol

Answer 1

The energy change when 249 mL of 15.0 M nitric acid is diluted with D.I. water is -25.83 kJ. The final temperature of the resulting solution is 19.2 °C.

Explanation

When diluting the 15.0 M nitric acid with water, an exothermic reaction occurs, releasing energy due to the decrease in concentration. Using the enthalpy change per mole of nitric acid (-6.88 kJ/mol) and the volume of the solution, the energy change can be calculated using the formula ΔH = n * ΔH°, where n is the number of moles. This yields an energy change of -25.83 kJ.

The temperature change resulting from mixing the acid with water can be calculated using the formula q = mcΔT, where q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Since the reaction releases energy, it heats the solution. To find the final temperature, the total heat released from the reaction is divided by the total mass of the solution times its specific heat capacity, giving the final temperature of 19.2 °C.

This calculation assumes ideal conditions, neglecting any heat loss to the surroundings and assuming the specific heat capacity of the resulting solution is close to that of water. It also assumes that the reaction between nitric acid and water is the only significant source of heat in this scenario.

Answer 2

The energy change of diluting a 15.0 M nitric acid solution by adding it to D.I. water is an exothermic reaction that releases 25.694 kJ of heat. This heat release leads to a decrease in temperature, resulting in a final temperature of approximately 10.6°C for the diluted solution.

Step-by-step explanation:

The student is asking about the energy change associated with diluting a concentrated nitric acid solution and the final temperature of the solution after dilution. Given that diluting 15.0 M nitric acid is an exothermic process (ΔH°= -6.88 kJ/mol), we can calculate the heat released during the dilution and then use this to find the final temperature of the resulting solution.Calculating the Moles of HNO3

First, calculate the moles of HNO3 using the initial volume and concentration:

Moles HNO3 = 15.0 M × 0.249 L = 3.735 moles

Calculating the Energy Change

Then, calculate the energy change:

Energy change (ΔH) = 3.735 moles × (-6.88 kJ/mol) = -25.694 kJ (negative sign indicates heat is released)

Calculating the Final Temperature

Assuming the heat capacity and density of the dilute solution resemble that of water and there's no heat loss, we use the following relationship:

Heat released by HNO3 = Heat absorbed by water

q = mcΔT, where m is the mass of water, c is the specific heat capacity of water (4.184 J/g°C), and ΔT is the change in temperature.

The mass of water (322 mL + 249 mL) ≈ 571 mL × 1 g/mL = 571 g

ΔT can be found by rearranging the equation:

ΔT = q / (mc) = -25.694 kJ / (571 g × 4.184 J/g°C) = -25.694 kJ / (2388.664 J/°C) = -10.757°C

Final Temperature = Initial Temperature + ΔT = 21.4°C - 10.757°C ≈ 10.6°C

Note that due to the heat being released, we subtract the temperature change to get the final temperature.