Question
The heat of solution of ammonium nitrate is 26.2 kJ/mol, and a 7.315 g sample of NH4NO3 is added to 40.0 mL of water in a constant pressure calorimeter initially at 25.2°C, what is the final temperature reached by the solution? (The specific heat of water = 4.184 J/g•°C)
Answers
Heat lost by KNO3 + heat gained by water = 0
q lost by NH4NO3 is
26.3 kJ x (7.315/80) = approx 2.5 kJ but you need a better answer.
Since dH is + you know the reaction is endothermic and the solution will get colder.
2.5 = mass H2O x specific heat H2O x delta T.
Solve for delta T and subtract from 25.2. Remember that 2.5 is approximate.
q lost by NH4NO3 is
26.3 kJ x (7.315/80) = approx 2.5 kJ but you need a better answer.
Since dH is + you know the reaction is endothermic and the solution will get colder.
2.5 = mass H2O x specific heat H2O x delta T.
Solve for delta T and subtract from 25.2. Remember that 2.5 is approximate.
Related Questions
ammonium nitrate decomposes to N2O and water at temperatures between 250 C and 300 C. Write a balanc...
When 10.0 g of ammonium nitrate (molar mass= 80 g/mol) are dissolved in 1.00 L of water in a bomb ca...
How many moles of ammonium nitrate is created from 10.7 grams of ammonium nitrate? Thus how many gra...
10.8 g of solid ammonium nitrate (NH4NO3, molar mass 80.052 g/mol) is dissolved in 150 mL of water....