Answer:
The experimental value of ΔH is -50 kJ/mol
Explanation:
<u>Step 1: </u>Data given
Volume of 1.0 M NaOH = 10.0 mL = 0.01 L
Volume of 1.0 M HCl = 10.0 mL = 0.01 L
Temperature before mixing = 20 °C
Final temperature = 26 °C
Specific heat of solution = 4.2 J/g°C
Density = 1g/mL
<u>Step 2: </u>Calculate q
q = m*c*ΔT
⇒ with m = the mass
⇒ 20.0 mL * 1g/mL = 20 grams
⇒ c = specific heat of solution = 4.2 J/g°C
⇒ ΔT = T2 -T1 = 26 -20 = 6 °C
q = 20g * 4.2 J/g°C * 6°C
q = 504 J
ΔHrxn = -q ( because it's an exothermic reaction)
ΔHrxn = -504 J
<u>Step 3:</u> Calculate number of moles
Moles = Molarity * volume
Moles = 1M *0.01 L = 0.01 moles
<u>Step 4:</u> Calculate the experimental value of ΔH
ΔHrxn = -504 / 0.01 mol = -50400 J/mol = -50.4 kJ/mol
The experimental value of ΔH is -50 kJ/mol
