NH4NO3( s )NH+4( a q )+NO3( a q ) In order to measure the enthalpy change for th
ID: 928773 • Letter: N
Question
NH4NO3(s)NH+4(aq)+NO3(aq)
In order to measure the enthalpy change for this reaction, 3.75 g of ammonium nitrate is dissolved in enough water to make 75.0 mL of solution. The initial temperature is 41.7C and the final temperature (after the solid dissolves) is 38.2 C.
Calculate the change in enthalpy for the reaction in kilojoules per mole. (Use 1.00g/mL as the density of the solution and 4.18J/gC as the specific heat capacity.)
I've attempted this several times and can not seem to get the correct answer, so any help is appreciated. Thanks!
Explanation / Answer
Hrxn = -q/n
n = mas/MW
MW = 80.052 of NH4NO3
n = 3.75/80.052 = 0.04684 mol of NH4NO3
Now,
Q = m*Cp*(tf-Ti)
m = m1+m2 = 3.75+75 = 78.75
Q = (78.75)(4.184)(38.2-41.7) = -1153.215 J
then
Hrxn = -(-1153.215)/(0.04684) = 24620.3 J
Hrxn = 24.6 kJ (is endothermic)
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