Question

I need helping with this pre-lab. Any help will be appreciated.

10A CHEMICAL EQUATIONS: ENERGY RELATIONSHIPS NAME 1315 Section Pre-Lab Assignment NOTE: The data used here are for illustrative purposes. The data you obtain will be different. A DETERMINATION OF CALORIMETER CONSTANT HCI (aq) + NaOH(aq) → NaCl (aq) + H2O (1) AH = -56.1 kJ/mol Record the molarity and volume of stock HCl solution used in this reaction and the initial, Ts, and final temperatures, T2. Report the temperature change for this reaction. EXAMPLE: used 25.0 mL of 2.31 M HCl; T, is 22.9°C; T2 is 35.4 °C; and AT is 12.5°C. PRACTICE: used 25.0 mL of 2.07 M HCl; T. is 24.3°C; Tis 35.9°C. Use the above data in conjunction with the given thermochemical equation to calculate the amount of heat produced in this reaction. EXAMPLE: -3.240 kJ. PRACTICE: Determine K. for the calorimeter. EXAMPLE: 0.259 kJ/°C PRACTICE: B. DETERMINATION OF AH FOR THE REACTION Mg(s) + 2HCl(aq) → MgCl2 (aq) + H2 (9) Record mass and temperature data, and report the temperature change for this reaction. EXAMPLE: the magnesium weighs 0.149 g; T; is 22.6°C; T2 is 36.4 °C; and AT is 13.8 °C. PRACTICE: the magnesium weighs 0.153 g; Tris 24.3 °C; Tis 38.5°C. Use the above results to determine the enthalpy change, AH, for this reaction. EXAMPLE: -583 kJ/mol Mg. PRACTICE:

Answer 1

A) The produced heat can be obtained by knowing that for each mol of reaction (1 mol of HCl or 1 mol of NaOH) the enthalpy is -56.1kJ/mol so, we need to know the number of moles of reaction and then multiply it by the molar enthalpy of reaction. First we'll obtain the number of moles of HCl:

$n_{reaction}=n_{HCl}=(25mL_{HCl})(\frac{2.07mol_{HCl}}{1,000mL_{HCl}})=0.0518mol$

Now we can multiply by the molar enthalpy:

$Q_{reaction}=(0.0518mol_{HCl})(-56.1*10^{3}J/mol)=-2905.98J$

As the heat produced by the reaction is the heat absorbed by the calorimeter (the calorimeter is an isolated system: there is not heat loss) we have:

$Q_{reaction}=-Q_{calorimeter}$

The heat absorbed by the calorimeter is:

$Q_{calorimeter}=K_c*\Delta T$

so:

$Q_{calorimeter}=K_c*\Delta T=2905.98J$

and isolating Kc:

$Kc=\frac{Q_{calorimeter}}{\Delta T}=\frac{2905.98J}{(35.9-24.3)^{o}C}=250J/^{o}C=0.250kJ/^{o}C$