Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin plastic divider. When the divider is broken, the ammonium nitrate dissolves according to the following endothermic reaction:

NH4NO3(s)→NH+4(aq)+NO−3(aq)
In order to measure the enthalpy change for this reaction, 1.25 g of NH4NO3 is dissolved in enough water to make 25.0 mL of solution. The initial temperature is 25.8 ∘C and the final temperature (after the solid dissolves) is 21.9 ∘C.

Calculate the change in enthalpy for the reaction in kilojoules per mole. (Use 1.0g/mL as the density of the solution and 4.18J/g⋅∘C as the specific heat capacity.)

After working it out I got -0.0064 kJ/mole, but it's not right. Anyone telling me how to solve this?

To solve this problem, we need to use the formula for calculating the change in enthalpy (ΔH) which is:

ΔH = q / n

where:
- ΔH is the change in enthalpy in J/mol
- q is the heat exchanged in J
- n is the amount of substance in mol

First, we need to calculate the heat exchanged (q) using the equation:

q = mcΔT

where:
- q is the heat exchanged in J
- m is the mass of the solution in g
- c is the specific heat capacity of the solution in J/g·°C
- ΔT is the change in temperature in °C

Let's calculate the mass of the solution first:

mass of solution = density × volume
mass of solution = (1.0 g/mL) × (25.0 mL) = 25.0 g

Now, we can calculate ΔT:

ΔT = final temperature - initial temperature
ΔT = 21.9 °C - 25.8 °C = -3.9 °C

Since the reaction is exothermic (heat is released), the final temperature is lower than the initial temperature, resulting in a negative ΔT.

Now, we can calculate q:

q = (25.0 g) × (4.18 J/g·°C) × (-3.9 °C)
q = -389.55 J

Next, we need to calculate the number of moles (n) of NH4NO3:

n = mass / molar mass
n = (1.25 g) / (80.0434 g/mol) = 0.0156 mol

Finally, we can calculate ΔH:

ΔH = -389.55 J / 0.0156 mol
ΔH = -25004.81 J/mol

To convert this to kJ/mol, we divide by 1000:

ΔH = -25004.81 J/mol / 1000 = -25.005 kJ/mol

Therefore, the change in enthalpy for the reaction is approximately -25.005 kJ/mol.

To solve this problem, we need to use the equation:

ΔH = q / n

where ΔH is the change in enthalpy, q is the heat transfer, and n is the number of moles of the substance involved in the reaction.

First, we need to calculate the heat transfer (q). To do this, we can use the equation:

q = m * Cp * ΔT

where q is the heat transfer, m is the mass of the solution, Cp is the specific heat capacity, and ΔT is the change in temperature.

1. Calculate the mass of the solution:
Since the density of the solution is given as 1.0 g/mL and the volume is 25.0 mL, the mass can be calculated as:
mass = density * volume = 1.0 g/mL * 25.0 mL = 25.0 g

2. Calculate the change in temperature:
ΔT = final temperature - initial temperature = 21.9 °C - 25.8 °C = -3.9 °C

Notice that the change in temperature is negative because the temperature decreased.

3. Plug in the values into the equation for q:
q = m * Cp * ΔT = 25.0 g * 4.18 J/g⋅∘C * -3.9 °C = -386.55 J

Next, we need to calculate the number of moles of NH4NO3 involved in the reaction.

4. Calculate the molar mass of NH4NO3:
NH4NO3 = 1(N) + 4(H) + 3(O) = 14.01 g/mol + 4(1.01 g/mol) + 3(16.00 g/mol) = 80.04 g/mol

5. Calculate the number of moles:
moles = mass / molar mass = 1.25 g / 80.04 g/mol = 0.0156 mol

Finally, we can calculate the change in enthalpy (ΔH) using the equation:

ΔH = q / n = -386.55 J / 0.0156 mol = -24,789.9 J/mol

Converting the units:

ΔH = -24,789.9 J/mol / 1000 J/kJ = -24.79 kJ/mol

Therefore, the change in enthalpy for the reaction is approximately -24.79 kJ/mol.

heat lost by warm water + heat gained by cool water = 0

[mass warm water x specific heat water x (Tfinal-Tinitial)] + [mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Solve for Tfinal.
Watch the units.