Molar enthalpy in calorimetry experiment

Question

25.0 mL of 1.00 M KOH solution is mixed with 25.0 mL of 1.05 M HCl in an open calorimeter. You can assume volumes of these solutions are additive. The density of the resulting solution is 1.03 g/mL and its heat capacity is 4.10 J/K/g. The temperature of the calorimeter and its contents rose by 6.70 °C. What is the molar enthalpy of reaction for KOH reacting with HCl? You can neglect the heat capacity of the calorimeter. 

Answer 1

This problem has several concepts.

First, list what you know: The equations fyou need to know are:

25.0 mL of 1.00 M KOH ∇H = q = mc∇T where m is mass, c is the heat capacity, and T is the

25.0 mL of 1.05 HCL temperature and ∇H is the enthalpy and q is the heat transferred.

Density = 1.03 g/mL n = CV where C is the concentration and V is the volume

c = 4.10 J/K/g

∇T = 6.70 °C.

Since enthalpy is equal to heat, we need to calculate q. First we need to calculate the mass from the density. Then we need to change the Temperature from Celsius to Kelvin to cancel out the unit Kelvin. This will be as follows:

m = 1.03 g/ml(50 ml) = 51.5 grams.

∇T = 6.70 + 273.15 = 279.85 K For this conversion, add 273.15 to the degrees. This gives the temp in K

Correction: the change in temperature would be 6.70 K. As J.R.S. pointed out, this is a change in the temperature. For whatever the initial and final temps are, the difference would still be 6.70 units.

q = mc∇T = 51.5 g(4.10J/K/g)(279.85 K) = 59.1kJ

correction: 51.5 g(4.10J/K/g)(6.7 K) = 1.41 X 10³ J

For convention, the value for the heat transferred is designated as positive or negative with respect to the system. Because in this system, the heat is leaving the system, the value of the heat will be negative. Therefore q = -1.41 X 10³ J.

For molar enthalpy of KOH, we need to find the number of moles:

KOH(moles) = CV = (1.0 mol/L)0.025 L) = 0.025 mol Don't forget to change mL to L

Then we divide the enthalpy by the number of moles:

-1.41 X 10³ J./0.025 mol = -56.6 kJ/mol

Note. This post has been updated. The text in bold is the corrections.