I need help on this chemistry work. Thanks. DATA ANALYSIS Using the example, cal

Question

I need help on this chemistry work. Thanks.

DATA ANALYSIS

Using the example, calculate the Gibbs free energy for the process of dissolving ammonium nitrate and the process of dissolving sodium hydroxide (equation: G = H – TS).

Use the standard temperature 298 K.

Use the experimental data to calculate enthalpy of dissolving (H).

Use the following data to calculate the entropy change (S):

S (J/K mol)

NH4NO3(s)

151

NH4+(aq)

113.4

NO3-(aq)

146.4            

NaOH(s)

64.46

Na+(aq)

59.0

OH-(aq)

- 10.7

After completing the calculations answer the following questions separately for each experiment:

Is the dissolving exothermic or endothermic?

Does the entropy decreases or increases during dissolving?

Is the Gibbs free energy positive or negative?   Is the process of dissolving spontaneous?

Using data in the appendix in your textbook, calculate Gibbs free energy for the reaction of dissolving

lithium phosphate, Li3PO4, in water at 25 °C. Do the calculations in two ways:

1) Use standard enthalpy of formation (H) and entropy (S) from the appendix in your textbook, and use the same formula as you did in the lab experiments: G = H – TS. Remember that both H and S has to include three lithium ions.

2) Use standard free energies of formation from Appendix 6 (G) and equation

G = G products – G reactants

Are the results similar in both calculations?

S (J/K mol)

NH4NO3(s)

151

NH4+(aq)

113.4

NO3-(aq)

146.4            

NaOH(s)

64.46

Na+(aq)

59.0

OH-(aq)

- 10.7

Explanation / Answer

I will help you with question 1, and the second, try to do it yourself:

You are not providing the values for dH but I manage to have some of them. In this case the dH for the compounds here are:

NH4NO3 = -366.1 kJ/mol
NaOH = -425.6 kJ/mol
NH4+ = -132.8 kJ/mol
NO3- = -206.56 kJ/mol
Na+ = 0
OH- = -229.94 kJ/mol

Now, let's calculate dH and dS for both:
H1 = (-132.8 -206.56) - (-366.1) = 26.74 kJ/mol
H2 = (-229.94) - (-425.6) = 195.66 kJ/mol

S1 = (146.3+113.4) - (151) = 108.8 J/mol K
S2 = (-10.7+59) - 64.46 = -16.16 J/mol K

Now th value of G:
G1 = 26740 - (298)(108.8) = -5682.4 J/mol or -5.68 kJ/mol
G2 = 195660 - (298)(-16.16) = 200,475.68 J/mol or simply 200.48 kJ/mol

Now, this values means that both reactions are not spontaneus (G is not equal to zero, to become spontaneus)

The sign of H will tell us if the reaction is exothermic or endothermic. As H > 1 means that both reactions are exothermic.

Following this example, try to do the other yourself.

Hope this helps

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