Question

In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter containing 75.0 g H2O initially at 23.2°C. The temperature of the water increased to 31.8°C. What was the change in enthalpy for the dissolution of the compound? Give your answer in units of joules per gram of compound. Assume that the specific heat of the solution is the same as that of pure water, 4.18 J ⁄ (g ⋅ °C).

Answer 1

The change in enthalpy for the dissolution of the compound is 344.11 J/g.

Step-by-step explanation:

In a coffee-cup calorimeter experiment, the change in enthalpy for the dissolution of a compound can be determined using the equation q = mCΔT, where q is the amount of heat absorbed or released, m is the mass of the compound, C is the specific heat of the solution, and ΔT is the change in temperature. Given that the initial temperature of the water is 23.2°C and the final temperature is 31.8°C, the change in temperature is ΔT = 31.8°C - 23.2°C = 8.6°C.

Since 10.00 g of the compound was added to 75.0 g of water, the total mass of the solution is 75.0 g + 10.00 g = 85.0 g.

Using the equation q = mCΔT, where C is the specific heat of water (4.18 J/(g·°C)), the amount of heat involved in the dissolution can be calculated as q = (85.0 g)(4.18 J/(g·°C))(8.6°C) = 3,441.14 J.

The change in enthalpy for the dissolution of the compound is thus 3,441.14 J/10.00 g = 344.11 J/g of the compound.

Answer 2

The answer is 269.6 J/g

Step-by-step explanation:

In a coffee-cup calorimeter, the heat released by the dissolution process of the ionic compound is absorbed by the water. So, heat of dissolution (
`q_(dis)`) is equal to the heat of water (
`q_(water)`).

We can first calculate the heat absorbed by the water:

`q_(water)`= m x Sh x ΔT

The specific heat of water(Sh) is known (4.18 J/g.ºC) and m is the mass of water in the calorimeter (75.0 g); ΔT is the change of temperature of water, from 23.2°C (initial temperature) to 31.8°C (final temperature). We introduce the data and calculate
`q_(water)`:

`q_(water)`= 75.0 g x 4.18 J/g.ºC x (31.8°C - 23.2°C)

`q_(water)`= 2696.1 J

As we know that heat of dissolution (
`q_(dis)`) is equal to the heat of water (
`q_(water)`) have already obtained
`q_(dis)`, and we have to only divide it into the mass of compound (10.00 g). In this way, we obtain the enthalpy change of dissolution in J per gram:

`q_(dis) =q_(water)`= ΔH/m= 2696.1 J/10.00 g= 269.61 J/g