Given: NH4NO3(s) + H2O(l) --> NH4OH (aq) + HNO3(aq) 50 mL of water Temperature c

ID: 1004258 • Letter: G

Question

Given: NH4NO3(s) + H2O(l) --> NH4OH (aq) + HNO3(aq)

50 mL of water

Temperature change: -3.8°C

Mass of NH4NO3 3.03g

a) Calculate the heat associated with water q(H2O) ____J

b) the heat either released or absorbed by the reaction(Change in Hrxn) _____J

c) Calculate the change in enthalpy (change in Hrxn) for this reaction _____kJ

d) Calculate the heat of reaction in kJ per mole of NH4NO3 ____kJ/mol

e) If the heat capacity of the calorimeter was 25J/ °C, what would the new change in enthalpy? ____kJ/mol

Given volume of water, V = 50 mL = 50 mL x (1 g / mL) = 50 g

Since the temperature change is negative, the final temperature is has decreased. Hence the reaction is endothermic.

(a):Heat associated with water, q(H2O) = mSdT = 50 g x (4.184 J/gxDecC) x (-3.8 DegC) = - 795 J (answer)

(b): Change in reaction, Hrxn = heat associated with water = 795 J (answer)

(c): Since the reaction is endothermic, change in enthalpy of reaction, Hrxn = + 795 J

(d): mass of NH4NO3 taken = 3.03 g

molar mass of NH4NO3 = 80.04 g/mol

Hence moles of NH4NO3 taken = mass/molar mass = 3.03 g / 80.04 g/mol = 0.037856 mol

Heat absorbed by 0.037856 mol NH4NO3 = + 795 J

=> Heat absorbed by 1 mol NH4NO3 = (+795 J / 0.037856 mol) x 1 mol = 21000 J/mol = 21 kJ/mol (answer)

(e): New heat change = 795 J + (25 J/DegC) x 3.8 DegC = 890 J

Enthalpy change = (+890 J / 0.037856 mol) x 1 mol = 23510 J/mol = 23.51 kJ/mol (answer)

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.