UNIT 6 REVIEW GAME

Endo/Exo

100

When 25.0 g of water absorbs heat and its temperature increases, is the process endothermic or exothermic?

Endothermic because the temperature increases

200

When 5.0 g of NH₄NO₃ dissolves in water, the solution temperature decreases from 22.0°C to 18.0°C. Is the dissolving process endothermic or exothermic?

Endothermic because the system gains heat and the temperature of the water decreases from 22 to 18.

300

A reaction is carried out in a coffee-cup calorimeter. The solution temperature increases from 21.5°C to 27.8°C. Is this process endothermic or exothermic?

Exothermic because the solution (surroundings) temperature increased, meaning the reaction released energy.

400

In a calorimetry experiment, a reaction absorbs 3.6 kJ of heat from the surroundings. What is the sign of ΔH, and is the reaction endothermic or exothermic?

Endothermic because the reaction absorbs heat from the surroundings.

H is positive (+3.6 kJ) because energy is going into the system.  

500

A reaction causes the temperature of water in a calorimeter to rise from 20.0°C to 26.5°C. Is the reaction endothermic or exothermic?

Exothermic because the water (surroundings) increases in temperature, meaning it gained heat from the reaction. So the reaction must be releasing heat.

Hess's law

100

C(s) + O2(g) yields CO2(g)        H=-393kJ

CO(g) + 1/2 O2(g) yields CO2(g)          H=-283

Find ΔH: 

C(s) + 1/2 O2(g) yields CO(g)

After reversing equation 2, the CO2 and 1/2 O2 cancel out. Then you add -393 and +283 it = -110

200

N2(g)+O2(g)→2NO(g)ΔH=+180 kJ 

2NO(g) + O2(g) → 2NO2(g)     ΔH=−114 

Find ΔH:

N2(g)+2O2(g)→2NO2(g)

when the equations are combined, the 2NO cancels out, and then you add the 180 -114 and get 66.

300

2S(s) + 3O2(g) → 2SO3(g)        ΔH=−594 

2S(s) + 2O2(g) → 2SO2(g)             ΔH=−594 

Find ΔH:

2SO2(g) + O2(g) → 2SO3(g)

You need to reverse equation 2 so that the 2S cancels out, and 2 of the 3 Oxygens on the left cancel out. Then you add -594 and 594 to end up with 0.

400

C(s) + O2(g) → CO2(g)              ΔH=−394 kJ

2CO(g) + O2(g) → 2CO2(g)            ΔH=−566kJ

Find ΔH:

2C(s) + O2(g) → 2CO(g)

First you need to multiply equation 1 by 2 and then reverse equation 2. You end up having to add -788 and +566 to get -222kJ

500

C(s) + O2(g) → CO2(g)               ΔH=−394 kJ

2CO(g) + O2(g) → 2CO2(g)            ΔH=−566 kJ

Find ΔH: 2C(s) + O2(g) → 2CO(g)

First, cancel out the O2 and 2CO2 by doubling reaction 1 and reversing reaction two. This leaves you having to add -788 and 566 to get -222kJ

Bond enthalpies

100

Use average bond enthalpies to estimate ΔH for this reaction:

H2(g)+Cl2(g)→2HCl(g)

Given:

• H–H = 436 kJ/mol
• Cl–Cl = 242 kJ/mol
• H–Cl = 431 kJ/mol

When shown the bonds the equation is broken - formed. there is 1 of both H2 and Cl2 so you add 436 +242 and there is 2 H-Cl bonds so you double 431 to get 862.Then you subtract 862 from (436+242) and get -84.

200

CH4(g)+2O2(g)→CO2(g)+2H2O(g)

Given:

• C–H = 413 kJ/mol
• O=O = 498 kJ/mol
• C=O (in CO₂) = 799 kJ/mol
• O–H = 463 kJ/mol

Broken - formed

413 x 4 = 1652, 498 x 2 = 996  996+1652= 2648

799 x 2 = 1598, 463 x 4=185   1852+1598=3450

2648 - 3450 = -802

300

Estimate ΔH for the reaction:

N2(g)+O2(g)→2NO(g)

Given bond enthalpies:

• N≡N = 945 kJ/mol
• O=O = 498 kJ/mol
• N=O = 607 kJ/mol

Broken - formed

945 + 498 = 1443

607 x 2 = 1214

1443-1214 = 229kJ

400

Estimate ΔH for the reaction:

C2H4(g)+H2(g)→C2H6(g)

Given bond enthalpies:

• C=C = 614 kJ/mol
• C–C = 347 kJ/mol
• C–H = 413 kJ/mol
• H–H = 436 kJ/mol

Broken - formed

614 + 4 x 413 + 436 = 2702

347 + 6 x 413 = 2825

2702 - 2825 = -123kJ

500

Estimate ΔH for the reaction:

HCN(g)+2H2(g)→CH3NH2(g)

Given bond enthalpies:

• C–H = 413 kJ/mol
• N–H = 391 kJ/mol
• C≡N = 891 kJ/mol
• C–N = 305 kJ/mol
• H–H = 436 kJ/mol

Broken - Formed

413 + 891 + 436 x 2 = 2176

(3 x 413) + (2 x 391) + 305 = 2326

2176 - 2326 = -150kJ

Heat capacity

100

A 50.0 g sample of water is heated from 20.0°C to 21.0°C.

How much heat was absorbed?

Given:

• c water=4.18 J/g C

q=mcΔT 

50.0 x 4.18 x 1.0 = 209J

200

A 50.0 g sample of water absorbs 840 J of heat. The temperature increases from 20.0°C to 24.0°C.

What is the specific heat capacity of the water?

q=mcΔT     c=q/(ΔTm)

840 / (50 x 4) = 4.2 J/g C

300

A 100.0 g sample of water at 30.0°C absorbs 2,090 J of heat.

What is the final temperature of the water?

Given:

• c water=4.18 J/g C

q=mcΔT        ΔT=q / mc

2090 / (100.0 x 4.18) = 5 C

400

A 50.0 g sample of a substance is heated from 18.0°C to 32.0°C. It absorbs 1,470 J of heat.

What is the specific heat capacity of the substance?

Given:

• q=1470 J
• m=50.0 g
• c=?

q=mcΔT         c = q/ mΔT 

1470 / (50.0 x 14.0) = 2.1 J/g C

500

A 200.0 g sample of water at 15.0°C absorbs 5,040 J of heat.

What is the final temperature of the water?

Given:

• c water = 4.18 J/g C

q=mcΔT      ΔT=q/mc 

5040 / (200.0×4.18) = 6.03

Initial temp + ΔT = Final temp

15.0 + 6.03 = 21.03