Consider the following Thermodynamic reactions, solve: A/ A 5.80 g sample of sol

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, H°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.

C/  Using the enthalpies of formation given below:

2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(l) H°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.

Equations: H°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

Explanation / Answer

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), H°rxn = - 56.5 KJ

H°rxn for the reverse of the above reaction is

NO2(g) ---> NO(g) + 1/2O2(g), H°rxn1 = - (- 56.5 KJ) = + 56.5 KJ ----(1)

Similarly reverse of the second reaction is

NO2(g) --- > 1/2N2(g) + O2(g) , H°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), H°rxn3 = +142.3 KJ /2 = 71.15 KJ ----(3)

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

----------------------------------------------------------------------------------------------------------------

(1) + (2) + (3) => 2NO2(g) ---> NO(g) + NO3(g), H°rxn = H°rxn1+H°rxn2+H°rxn3

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

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