u16q diagnostic KEY

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Score = /22 = %

Chapter 16 Diagnostic Name:

KEY

1. 20 mL of 0.1 M H

3

PO

4

and 20 mL of 0.3 M NaOH are combined in a coffee cup calorimeter. The temperature of the solution increases by 1.90°C when the following reaction occurs.

H

3

PO

4

(aq) + 3NaOH(aq)  Na

3

PO

4

(aq) + 3H

2

O(l)

The specific heat of the solution is 4.18. What is the ΔH of the reaction? (6 pts)

First we find the heat gained by the solution in kilojoules: q = mcΔT = 40(4.18)(1.90) = 317.7 J = 0.3177 kJ

We must also know number of moles that reacted:

20 mL = 0.020 L

Then calculate ΔH:

0.020(0.1) = 0.002 mol of H q = nΔH

-0.3177 = 0.002(ΔH)

ΔH = -158.8 kJ/mol

2. Find the ΔH for the reaction below. (6 pts)

(g)

3

PO

4

CH

4

(g) + NH

3

(g)  HCN(g) + 3H

2

Use the following reactions and their ΔH values:

N

2

(g) + 3H

2

(g) → 2NH

3

(g) ∆H = -91.8 kJ/mol

C(s) + 2H

2

(g) → CH

4

(g) ∆H = -74.9 kJ/mol

H

2

(g) + 2C(s) + N

2

(g) → 2HCN(g) ∆H = +270.3 kJ/mol

CH

4

(g)  C(s) + 2H

2

(g)

NH

3

(g)  1/2 N

2

(g) + 3/2 H

1/2 H

2

(g) + C(s) + 1/2 N

2

(g) → HCN(g)

2

∆H = +74.9

(g) ∆H = +91.8/2

∆H = +270.3/2

ΔH = +255.95 kJ/mol

3. Given the following Lewis structures and bond enthalpy values:

Bond Bond Enthalpy (kJ/mol) Bond Bond Enthalpy (kJ/mol)

C-C 347

C=C 598

C  C 813

C-N 285

C=N 616

C-H 413

C=O 805

O=O 498

H-O

H-Cl

464

431

C  N 866

C-Br 285

N=N 418

N  N 946

H-N

H-H

391

436

Br-Br 193

Cl-Cl 242

N-O 222 N=O 590 a.

Calculate the ΔH of the following reaction: (6 pts)

CH

4

(g) + NH

3

(g)  HCN(g) + 3H

2

(g)

Sum of the bond enthalpies of the reactants (energy in) = 4(413) + 3(391) = 2825

Sum of the bond enthalpies of the products (energy out)= 413 + 866 + 3(436) = 2587

ΔH = 2825 – 2587 = +238 kJ/mol b.

For the reaction in part a, would the ΔS be positive or negative? Explain your answer.

(2 pts)

Positive. The products consist of 4 moles of gas, the reactants only 2. When more gas is produced, entropy increases. c.

Based on your answers to parts a and b, would you expect the reaction to be: (2 pts) a.

always spontaneous b.

never spontaneous c.

spontaneous if the temperature is high enough d.

spontaneous if the temperature is low enough

Explain your answer.

The positive ΔH favors a nonspontaneous reaction, but the positive ΔS favors a spontaneous reaction. To make the TΔS term outweigh the ΔH, the temperature must be high.

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