When pure sulphuric acid dissolves in water, large amount of heat is given off. To measure it, 175 g of water was placed in a coffee cup calorimeter and chilled to 1 0 ∘ C . Then 4.9 g of sulphuric ac

Question

When pure sulphuric acid dissolves in water, large amount of heat is given off. To measure it, 175 g of water was placed in a coffee cup calorimeter and chilled to 1 0 ∘ C . Then 4.9 g of sulphuric acid ( H 2 ​ SO 4 ​ ) , also at 1 0 ∘ C , was added, and the mixture was quickly stirred with a thermometer. The temperature rose rapidly to 14. 9 ∘ C . Since the solution is dilute, assume that the value of the specific heat of the solution is 4.184 J g − 1 ∘ C − 1 (same as that of water), and that the solution absorbs all the heat evolved. Calculate the heat evolved per mole of sulphuric acid (in kJ ) (Assume work due to stirring is negligible)

TestHub · Accepted Answer

The heat released in the process is given by

$q =$ Mass × Specific heat × $Δ T$

$= 4.184 \times (175 + 4.9) \times (14.9 - 10)$

$= 3.7 \times 1 0^{3} J = 3.7 kJ$

This should then be converted to a value representing kJ per mole of reactant, remembering that the sign of $Δ H$ is to be negative, since the process releases heat energy to surroundings. The number of moles of sulphuric acid is : $n = 4.90 \times (\frac{1}{98.00}) = 5.00 \times 1 0^{- 2} mol$

and the enthalpy change per mole is given by

$Δ H = \frac{3.7}{0.05} = 74 kJ mol^{- 1}$

Chemistry - Thermodynamics - I Question with Solution | TestHub

Doubts & Discussion

JEE Advanced (2028)