e pr Problem 5.56 Enhanced with Feedback Part A When a 4.25-g sample of solid am
ID: 549109 • Letter: E
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e pr Problem 5.56 Enhanced with Feedback Part A When a 4.25-g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter, the temperature drops from 22.0 C to 16.9 °C Calculate H (in kJ/mol NH4NO3) for the solution process NH, NO3(a)-NH: (aq) + NO3(aq) Assume that the specific heat of the solution is the same as that of pure water Express your answer to two significant figures and include the appropriate units. stirer ? Cork stopper Value Units Submit My Answers Give Up Two Styrofoum® cups nested together Reaction mixture in soluton Part B Is this process endothermic or exothermic? You may want to reference (n pages 179 183) Section 5.5 while completing this problem. endoth exothermic Sulrnit My Answers Give UpExplanation / Answer
Given reaction: NH4NO3(s) -------> NH4+(aq) + NO3-(aq)
Since it is an isolated closed system, the total energy of the system is zero.
q(reaction) + q(surroundings) = 0
q = m. Cp. T
m = mass of the substance
Cp = specific heat (j/g.K)
T = Change in temperature.
Let's calculate the heat gained or lost by the surroundings:
we have, m = 60 g + 4.25 g = 64.25 g
Cp = 4.184 J/g.K (specific heat of pure water)
T = T2-T1 = 16.9-22 = - 5.1 oC
qsurroundings = 64.25 g x 4.184 J/g.K x (- 5.1 oC)
= - 1371 J
Therefore, qrxn = -qsurr = -(-1371 J) = 1371 J
hence, the enthalpy change of the reaction at constant pressure is;
H = qrxn = 1371 J = 1.371 KJ
Therefore, 4.25 g of NH4NO3 requires 1.371 KJ of energy. Now, we need to calculate the H per one mole of NH4NO3.
no. of moles of NH4NO3 = mass/mol.mass = 4.25/80 g/mol = 0.0531 moles.
0.0531 moles. of NH4NO3 requires ------------1.371 KJ
for 1 mole NH4NO3 requires------------------------------?
H = 1.371 KJ/0.0531 moles = 25.82 KJ/mol
Hence, the answer is H = 26 KJ/mol (two significant numbers)
PartB: Answer: Endothermic reaction.
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