Question

When 26.1 mL of 0.500 M H2SO4 is added to 26.1 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.) (d for water = 1.00 g/mL; c for water = 4.184 J/g·°C.)

Answer 1

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