posted by DeeDee .
NH4NO3(s) NH4+(aq) + NO3−(aq)
In order to measure the enthalpy change for this reaction above, 1.07 g of NH4NO3 is dissolved in enough water to make 25.0 mL of solution. The initial temperature is 25.8°C and the final temperature (after the solid dissolves) is 22.4°C. Calculate the change in enthalpy for the reaction in kJ. (Use 1.0 g/mL as the density of the solution and 4.18 J/g · °C as the specific heat capacity.)
We've been doing a lot of energy problems, but I'm not sure how to begin this one. I don't want the answer, just the set up so I know what I'm doing. Thanks. :)
The rise is T for H2O is your route to heat involved.
q = mass H2O x specific heat H2O x delta T.
Then q/1.07 = delta H/gram.
Usually these are quoted in kJ/mol.
(delta H/gram NH4NO3) x molar mass NH4NO3 x (1 kJ/1000 J) = ?
Note: Since T went down you know it was cooled and means an endothermic reaction so the sign of delta H is +.
Would the answer be 27kJ?
I think you can report more than 2 significant figures.
I calculated 26.6 kJ/mol.