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When 23.6 mL of 0.500 M H2SO4 is added to 23.6 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.)

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Answer:

-112 Kj/mole (diprotic acid)

Explanation:

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The enthalpy of reaction is -55 KJ/mol.

We must first write down the equation of the reaction;

2KOH(aq) + H2SO4(aq) ------> K2SO4(aq) + 2H2O(l)

Then we compute the number of moles of H2SO4 = 23.6/1000 × 0.500 M  = 0.012 moles

And the number of moles of KOH = 23.6/1000 × 1.00 M = 0.024 moles

From what we know in the reaction equation;

2 moles of KOH produces 2 moles of water

Therefore, 0.0026 moles of KOH produces 0.024 moles of water.

The total volume of solution is obtained by adding  = 23.6 mL + 23.6 mL = 47.2 mL

Mass of water = density × volume = 1.00 g/mL ×  47.2 mL =47.2 g

Using the formula;

ΔH = mcθ

Mass of solution (m) = 47.2 g

Specific heat capacity of solution (c) =  4.184 J/g·°C

Temperature difference(θ) =  30.17°C - 23.50°C = 6.67°C

Substituting values;

ΔH = -( 47.2 g × 4.184 J/g·°C  × 6.67°C)/ 0.024 moles  

ΔH =  -(1.32 KJ/0.024 moles)

ΔH = -55 KJ/mol

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