When 1.477 mole of an inorganic salt were dissolved in 1.00 L of water in a constant pressure solution calorimeter, the temperature of the calorimeter changed from 25.00oC to 24.40 oC. The total heat capacity of the calorimeter including all its contents was 2.18 kJ oC-1
What is the molar enthalpy of solution for this salt, DH / J mol-1?
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To find the molar enthalpy of solution (ΔH) for the salt, we can use the equation:
ΔH = q / n,
where ΔH is the molar enthalpy of solution, q is the heat exchanged in the process, and n is the number of moles of the salt.
In this scenario, the heat exchanged (q) can be calculated using the equation:
q = CΔT,
where q is the heat exchanged, C is the heat capacity of the calorimeter, and ΔT is the change in temperature.
Given that the total heat capacity of the calorimeter (including all its contents) is 2.18 kJ oC-1, we need to convert this to J oC-1 by multiplying by 1000:
C = 2.18 kJ oC-1 × 1000 J kJ-1 = 2180 J oC-1.
The change in temperature (ΔT) is calculated by subtracting the initial temperature from the final temperature:
ΔT = 24.40 oC - 25.00 oC = -0.60 oC.
Next, we need to calculate the heat exchanged (q) using the equation:
q = CΔT.
q = 2180 J oC-1 × -0.60 oC = -1308 J.
Now, we can calculate the molar enthalpy of the solution (ΔH) using the equation:
ΔH = q / n.
Given that there are 1.477 moles of the salt, we can substitute the values into the equation:
ΔH = -1308 J / 1.477 mol = -885 J mol-1.
Therefore, the molar enthalpy of solution (ΔH) for this salt is -885 J mol-1.