Question

Consider the following Thermodynamic reactions, solve:

A/ A 5.80 g sample of solid NH4Br (s) is dissolved in 117 mL of water in a coffee cup calorimeter. Once all of the NH4Br (s) is dissolved in the water, the final temperature of the solution is found to be 6.73°C. If the initial temperature of the water in the calorimeter was 21.63 °C, calculate the calorimeter constant (in J/K) for the coffee cup calorimeter. Report your answer to three significant figures. The heat of solvation of NH4Br (s) is 16.78 kJ/mol.

B/ Calculate the standard enthalpy change, ΔH°rxn, in kJ for the following chemical equation, using only the thermochemical equations below:

2NO₂(g) → NO(g) + NO₃(g)
Report your answer to three significant figures in scientific notation.

Equations:		ΔH°rxn (kJ)
NO(g) + 1/2O2(g) → NO2(g)		-56.5
1/2N2(g) + O2(g) → NO2(g)		33.8
N2(g) + 3O2(g) → 2NO3(g)		142.3

C/  Using the enthalpies of formation given below:

2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(l) ΔH°_rxn = -1124.14

H₂S (g): -20.60 kJ/mol

O₂ (g): 0.00 kJ/mol

SO₂ (g): -296.84 kJ/mol

H₂O (l): -285.83 kJ/mol

Calculate the amount of heat absorbed/released (in kJ) when 8.39 grams of SO₂ are produced via the above reaction.

Report your answer to two decimal places, and use appropriate signs to indicate heat flow direction.

Answer 1

Q.1: Given the mass of solid NH4Br (s) taken = 5.80 g

Molar mass of NH4Br (s) = 97.94 g/mol

Hence moles of solid NH4Br (s) taken = 5.80 g / 97.94 g/mol = 0.0592 mol

Since the temperature of calorimeter decreases, energy is absorbed during solvation.

Given the solvation energy of NH4Br (s) = 16.78 kJ/mol.

Hence energy absorbed during the dissolution of 5.80 g ( = 0.0592 mol) of NH4Br (s)

= 16.78 kJ/mol. x 0.0592 mol

= 0.993376 KJ = 0.993376 KJ x (1000J/1KJ) = 993.376 J

Temperature change, dT = 6.73 DegC - 21.63 DegC = - 14.9 DegC

Let the  calorimeter constant (in J/K) be C J/K

Hence heat released by calorimeter = C x dT

According law of calorimetry

Hence heat released by calorimeter = - ( energy absorbed during the dissolution)

=> C x dT = - 993.376 J

=> C x ( - 14.9 DegC) = - 993.376 J

=> C = 993.376 J / ( - 14.9 DegC) = 66.7 J/K (answer)

Q.2:Given

NO(g) + 1/2O₂(g) ---> NO₂(g), ΔH°_rxn = - 56.5 KJ

ΔH°_rxn for the reverse of the above reaction is

NO₂(g) ---> NO(g) + 1/2O₂(g), ΔH°_rxn1 = - (- 56.5 KJ) = + 56.5 KJ ----(1)

Similarly reverse of the second reaction is

NO₂(g) --- > 1/2N₂(g) + O₂(g) , ΔH°_rxn2 = - (+33.8 KJ) = - 33.8 KJ ----(2)

On dividing the 3rd reaction by 2 we get

1/2N₂(g) + 3/2O₂(g) ---> NO₃(g), ΔH°_rxn3 = +142.3 KJ /2 = 71.15 KJ ----(3)

Now we will the desired reaction by adding reaction (1), (2) and (3)

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(1) + (2) + (3) => 2NO₂(g) ---> NO(g) + NO₃(g), ΔH°_rxn = ΔH°_rxn1+ΔH°_rxn2+ΔH°_rxn3

=> ΔH°_rxn = + 56.5 KJ + ( - 33.8 KJ ) +  71.15 KJ = 93.85 KJ = 9.38 x 10¹ KJ (answer)