Question

When sulfuric acid dissolves in water, a great deal of heat is given off. The enthalpy change for this process is called the enthalpy of solution. To measure it, 175 g of water was placed in a coffee-cup calorimeter and chilled to 10°C. Then 49.0 g of pure sulfuric acid, also at 10.0°C, was added, and the mixture was quickly stirred with a thermometer. The temperature rose rapidly to 14.9°C. Assume that the value of the specific heat of solution is 4.184 J/g°C. You may assume that the specific heat of the resulting sulfuric acid solution will also be 4.184 J/g°C. Which of the following provides the enthalpy change in kilojoules per mole of H2SO4?

a) -53.57 kJ/mol
b) -45.62 kJ/mol
c) -36.88 kJ/mol
d) -64.74 kJ/mol

Answer 1

The enthalpy change when sulfuric acid dissolves in water was calculated to be 9.17 kJ/mol, but this does not match any of the provided answer options, suggesting there may be an error in the given choices or the calculation.

Step-by-step explanation:

To determine the enthalpy change in kilojoules per mole of H2SO4 when sulfuric acid dissolves in water, we need to perform a few calculations based on the provided data:

• 
  
• 
  
• 
  

Plugging in our values:

q = 224.0 g × 4.184 J/g°C × 4.9°C = 4584.75 J

Now, we convert joules to kilojoules (since 1 kJ = 1000 J):

q = 4584.75 J ÷ 1000 J/kJ = 4.585 kJ

The molar mass of H2SO4 is approximately 98.08 g/mol, which means the number of moles of H2SO4 in 49.0 g is:

moles of H2SO4 = 49.0 g ÷ 98.08 g/mol = 0.5 mol

To find the enthalpy change per mole, we divide the heat (q) by the number of moles (n) of H2SO4:

ΔH = q ÷ n = 4.585 kJ ÷ 0.5 mol = 9.17 kJ/mol

Note that this is a negative value because the process is exothermic, meaning heat is released.

The closest answer choice to our calculation is (b) -45.62 kJ/mol, which suggests there may have been a mistake either in the calculation or in the answer choices given. For a more accurate assessment, we'd need to verify all values and calculations.