Answer:
ΔH = -55816 J / mol = -55.8 kJ/mol
Explanation:
Step 1: data given
Volume of a 0.500 M H2SO4 solution = 26.1 mL = 0.0261 L
Volume of a 1.00 M KOH = 26.1 mL = 0.0261 L
Temperature of the coffee cup = 23.50° C
The temperature rises to 30.17°C
Step 2: The balanced equation
H2SO4 + 2KOH → K2SO4 + 2H2O
Step 3: Calculate moles
Moles = molarity * volume
Moles H2SO4 = 0.500 M * 0.0261 L
Moles H2SO4 = 0.01305 moles
Moles KOH =1.00 M * 0.0261 L
Moles KOH = 0.0261 moles
Step 4: Calculate the limiting reactant
The mol ration is 1:2
So H2SO4 and KOH is both completely used.
Step 5: Calculate total volume
Total volume = 26.1 mL + 26.1 mL = 52.2 mL
Step 5: Calculate mass
Mass = density * volume
Mass = 52.2 mL * 1.00 g/mL
Mass = 52.2 grams
Step 6: Calculate Q
Q = c*ΔT*m
⇒c = the specific heat of water = 4.184 J/g°C
⇒ΔT = the change of temperature = 30.17 - 23.50 = 6.67 °C
⇒m = the mass = 52.2 grams
Q = 4.184 * 6.67 * 52.2
Q = 1456.8 J
As, heat change of reaction = -(Heat change of solution)
Therefore, heat change of reaction = -1456.8 J
Step 7: Calculate ΔH
There will be produced 0.0261 moles H2O
ΔH = Q / n
ΔH = -1456.8 J / 0.0261 moles
ΔH = -55816 J / mol = -55.8 kJ/mol
