there's a 5.44 g sample of NH4NO3 and 150 g sample of water. So they ask for the change in enthalpy in kJ/mol. They give the temperature change from 16.2 degrees C to 18.6 degrees C. They also provide the specific heat capacity of the entire solution (155.4g) as 4.2 J/g*K. How do you solve for the change in enthalpy?? I kept getting the wrong answer. I thought you had to just use the equation q = mc*(change in temperature)

To solve for the change in enthalpy (∆H) of the NH4NO3 solution, you need to consider the heat change that occurred during the temperature change.

The equation you mentioned, q = mc*∆T, can be used to calculate the heat flow (q) when a mass (m) of a substance with a specific heat capacity (c) undergoes a temperature change (∆T). However, in this case, you need to consider the heat flow for the entire solution, which consists of both NH4NO3 and water.

To solve this problem correctly, you should consider the individual heat flows for both the NH4NO3 and the water, and then sum them up to find the total heat flow (∆H).

Here's a step-by-step approach to solve for the change in enthalpy:

1. Calculate the heat flow for the NH4NO3:
q_NH4NO3 = m_NH4NO3 * c_NH4NO3 * ∆T

- Given: m_NH4NO3 = 5.44 g (mass of NH4NO3), c_NH4NO3 = ? (specific heat capacity of NH4NO3), ∆T = 18.6°C - 16.2°C

- The specific heat capacity of NH4NO3 is not provided. You can find it in a table or look it up online. Let's assume it to be 2.0 J/g*K for the calculation.

q_NH4NO3 = 5.44 g * 2.0 J/g*K * (18.6°C - 16.2°C)

2. Calculate the heat flow for the water:
q_water = m_water * c_water * ∆T

- Given: m_water = 150 g (mass of water), c_water = 4.2 J/g*K (specific heat capacity of water), ∆T = 18.6°C - 16.2°C

q_water = 150 g * 4.2 J/g*K * (18.6°C - 16.2°C)

3. Calculate the total heat flow:
q_total = q_NH4NO3 + q_water

q_total = q_NH4NO3 + q_water

4. Convert the total heat flow to kJ/mol:
∆H = q_total / n

- The number of moles (n) is calculated using the molar mass of NH4NO3. The molar mass of NH4NO3 can be found by adding up the atomic masses of its constituent elements (N + H + H + H + O + O + O).

∆H = q_total / n

Once you have these calculations, you'll have the change in enthalpy (∆H) in kJ/mol. Make sure to use the correct values for the specific heat capacities and carefully perform the calculations to obtain an accurate result.

To solve for the change in enthalpy (ΔH), you need to consider the heat absorbed or released by both the NH4NO3 and water, and then calculate the overall change in enthalpy.

1. Calculate the heat absorbed or released by NH4NO3:
- Convert the mass of NH4NO3 to moles using its molar mass.
- Multiply the moles of NH4NO3 by the molar enthalpy of NH4NO3 (ΔH_NH4NO3). The ΔH_NH4NO3 for dissolution is -25.7 kJ/mol.

2. Calculate the heat absorbed or released by water:
- Convert the mass of water to moles by dividing by the molar mass of water. The molar mass of water is 18.015 g/mol.
- Multiply the moles of water by the specific heat capacity of water (4.2 J/g•K) and the change in temperature (ΔT).

3. Sum up the heat absorbed or released by NH4NO3 and water to find the overall change in enthalpy (ΔH).

Here are the step-by-step calculations:

1. Calculate the heat absorbed or released by NH4NO3:
- Mass of NH4NO3 = 5.44 g
- Molar mass of NH4NO3 = 80.0434 g/mol
- Moles of NH4NO3 = 5.44 g / 80.0434 g/mol = 0.06796 mol
- ΔH_NH4NO3 = -25.7 kJ/mol
- Heat absorbed or released by NH4NO3 = 0.06796 mol * (-25.7 kJ/mol) = -1.746 kJ

2. Calculate the heat absorbed or released by water:
- Mass of water = 150 g
- Molar mass of water = 18.015 g/mol
- Moles of water = 150 g / 18.015 g/mol = 8.328 mol
- Specific heat capacity of water = 4.2 J/g•K
- Change in temperature (ΔT) = 18.6°C - 16.2°C = 2.4°C = 2.4 K
- Heat absorbed or released by water = 8.328 mol * 4.2 J/g•K * 2.4 K = 80.026 J

3. Calculate the overall change in enthalpy (ΔH):
- ΔH = Heat absorbed or released by NH4NO3 + Heat absorbed or released by water
- ΔH = -1.746 kJ + 80.026 J * (1 kJ / 1000 J)
- ΔH = -1.746 kJ + 0.080026 kJ = -1.666 kJ (rounded to three decimal places)

Therefore, the change in enthalpy (ΔH) for the given reaction is approximately -1.666 kJ/mol.