Assume you mix 100.0 mL of 200 M CsOH with 50.0 mL 0f 0.400 M HCl in a coffee cup calorimeter. A reaction occurs. The temperature of both solutions before mixing was 22.50 °C, and it rises to 24.28 °C after the acid-base reaction. What is the enthalpy change for the reaction per mole of CaOH? Assume densities of the solutions are all 1.00 g/mL and the specific heat capacities of the solutions are 4.2 J/gK

The enthalpy change for the reaction per mole of CsOH is -56.1 kJ/mol

Explanation:

Step 1: Data given

Volume of a CsOH solution = 100.0 mL

Molarity of a CsOH solution = 0.200 M

Volume of HCl solution = 50.0 mL

Molarity of HCl solution = 0.400M

The temperature of both solutions before mixing was 22.50 °C, and it rises to 24.28 °C after the acid-base reaction.

Density = 1.00 g/mL

Specific heat = 4.2 J/gK = 4.2 J/g°C

Step 2: The balanced equation

CsOH + HCl → CsCl + H2O

Step 3: Calculate the energy

Q = m*c*ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with m = the mass of the solution = (100+ 50 mL) * 1.00 g/mL = 150 grams

⇒with c = the specific heat of the solution = 4.2 J/g°C

⇒with ΔT = The change of temperature = T2 - T1  = 24.28 °C - 22.50 °C = 1.78 °C

Q = 150 grams * 4.2 J/g°C * 1.78 °C

Q = 1121.4 J

Step 4: Calculate moles CsOH

Moles CsOH = molarity CsOH * volume CsOH

Moles CsOH = 0.200M * 0.100 L

Moles CsOH = 0.0200 moles

Step 5: Calculate  the enthalpy change for the reaction per mole of CsOH

ΔH is negative since this is an exothermic reaction

ΔH = -Q/moles

ΔH = -1121.4 J / 0.0200 moles

ΔH = -56070 J/mol = -56.1 kJ/mol

The enthalpy change for the reaction per mole of CsOH is -56.1 kJ/mol