Answer:
The heat of the reaction = -1985 J = -1.985 kJ
The enthalpy of the reaction is -39.7 kJ/ mol
Explanation:
<u>Step 1: </u>Data given
Volume of HCl = 100 mL the heat of the reaction = 0.1 L
Molarity of HCl solution = 0.500 M
Volume of KOH = 100 mL = 0.1 L
Molarity of KOH solution = 0.500 M
Initial temperature = 23.0 °C
Final temperature = 25.5 °C
Specific heat of the solution = 3.97 J/°C *g
Density of the solution = 1g/ mL
<u>Step 2: </u>Calculate heat
Q = m*c*ΔT
with m = the mass of both solution : 100g + 100 g ( since density = 1g/mL) = 200 g
c = the specific heat = 3.97 J/°C*g
ΔT = T2 -T1 = 25.5 = 23 = 2.5 °C
Q = 200g *3.97 J/°C*g * 2.5°C = 1985 J (= -1985 J because it's exothermic)
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<u>Step 3:</u> Calculate the number of moles
Number of moles = Molarity * Volume
Number of moles = 0.5 * 0.1 L = 0.05 moles
(Moles of the acid are equal to the moles of water produced.
Moles of solution = 0.05 moles)
<u>Step 4: </u>Calculate the enthalpy of the reaction
ΔH = heat change /Number of moles
= -1.985 kJ/ 0.05 moles
=- 39700 J/mol = -39.7 kJ/ mol
The enthalpy of the reaction is -39.7 kJ/ mol
The enthalpy of the reaction is negative, this means the reaction is exothermic ( which means the final temperature is higher than the initial temperature.)
