thermo-chemistry enthalpy changes in chemical process part 1

Faculty-Wide Courses 2009-2010 Thermo-Chemistry Enthalpy Changes in Chemical Process

Eng. ISLAM IBRAHIM FEKRY Mob. 0109790568 Mail. [email protected]

SHEET No.2

PART 1

1. Two forms of carbon are graphite, the soft, black, slippery material used in “lead”

pencils and a lubricant for locks, and diamond, the brilliant, hard, gemstone. Using

the enthalpies of combustion for graphite (-394 KJ/mol) and diamond (-396

KJ/mol), Calculate Δ H for the conversion of graphite to diamond:

C graphite (solid) C diamond (solid)

Given that

Cgraphite(s) + O2(g) CO2(g) ΔH = - 394 kJ/mole

Cdiamond(s) + O2(g) CO2(g) ΔH = - 396 kJ/mole

To solve this problem we simply switch the direction of the 2nd reaction (the diamond reaction) and add the two reactions together. That is,

Cgraphite(s) + O2(g) CO2(g) ΔH1 = - 394 kJ/mole

CO2(s) Cdiamond(s) + O2(g) ΔH2= + 396 kJ/mole

_________________________________________ _____________________

Cgraphite(s) Cdiamond(s) ΔHtotal = ΔH1 + ΔH2 = 2 kJ/mole.



2. Methanol (CH3OH) is often used as a fuel in high performance engines in race cars.

Compare the standard enthalpy of combustion per gram of methanol with that per

gram of gasoline. A gasoline is actually a mixture of compounds, but assumes for

this problem that gasoline is a pure liquid octane (C8H18). (Standard enthalpies of

formation of CH3OH , C8H18 , CO2(g) and H2O(L) are

-239, -269, -394, and -286 KJ/mol respectively )

standard enthalpy of combustion per gram of methanol

CH3OH + 3

2 O2 CO2 + 2 H2O

Δ HC= ∑ Vi * Δ HF i

Δ HC CH3OH = (1 * Δ HF CO2 ) + (2 * Δ HF H20 ) + (-1 * Δ HF CH3OH ) + (- 3

2 * Δ HF O2 )

Δ HC CH3OH = (1 * -394 ) + (2 * -286 ) + (-1 * -239 ) + (-3

2 * 0 ) = - 727 KJ/mole

Mass CH3OH = NO. Mole CH3OH * M.WT CH3OH

Mass CH3OH = 1 * (12+4+16) = 1*32 = 32 gm


= Δ HC CH 3OH

M.wt CH 3OH =

−727

32 = -22.7 KJ/gm

standard enthalpy of combustion per gram of octane

C8H18 + 25

2 O2 8 CO2 + 9 H2O


Δ HC C8H18 = (8 * Δ HF CO2 ) + (9 * Δ HF H20 ) + (-1 * Δ HF C8H18 ) + (- 25

2 * Δ HF O2 )

Δ HC C8H18 = (8 * -394 ) + (9 * -286 ) + (-1 * -269 ) + (- 25

2 * 0 )

Δ HC C8H18 = ( - 3,152 ) + ( - 2,574 ) + ( 269 ) + ( 0 ) = - 5,457 KJ/mole

Mass C8H18 = NO. Mole C8H18 * M.WT C8H18

Mass C8H18 = 1 * ( 8*12+1*18) = 114 gm


= Δ HC C8H18

M.wt C8H18 =

−5,457

114 = - 47.86 KJ/gm

standard enthalpy of combustion per gram of octane > standard enthalpy of combustion

per gram of methanol



3. Compare that energy available from the combustion of a given volume of methane

and the same volume of hydrogen at the same temperature and pressure. (standard

enthalpy of formation of methane is -75 KJ/mol )

Since, they (methane and hydrogen) have the same volume and operated at same

temperature and pressure, So that they have the same No. of moles.

a-The combustion of methane

CH4 + 2 O2 CO2 + 2 H2O


Δ HC CH4 = (1 * Δ HF CO2 ) + (2 * Δ HF H20 ) + (-1 * Δ HF CH4 ) + (-2 * Δ HF O2 )

Δ HC CH4 = (1 * -394 ) + (2 * -286 ) + (-1 * - 75 ) + (-2 * 0 )

Δ HC CH4 = (-394 ) + ( -572 ) + (75 ) + (0 ) = - 891 KJ/mole

b-The combustion of Hydrogen

H2 + 1

2 O2 H2O

Δ HC H2 = (1 * Δ HF H20 ) + (-1 * Δ HF H2) + (- 1

2 * Δ HF O2 )

Δ HC H2 = (1 * -286 ) + (-1 * 0 ) + (- 25

2 * 0 )

Δ HC C8H18 = ( - 286 ) + ( 0) + ( 0 ) = - 286 KJ/mole

standard enthalpy of combustion of methane > standard enthalpy of combustion hydrogen



4. Assuming that the combustion of hydrogen gas provide three times as much energy

per gram as gasoline, calculate the volume of liquid H2 (Density = 0.0710 g/mL )

required to furnish the energy contained in 80.0 L of gasoline ( Density = 0.740

g/mL ). Calculate also the volume that this hydrogen would occupy as a gas at 1.00

atm and 25𝑜C.



5. When 1 mole of methane (CH4) is burned at a constant pressure, 890 KJ of energy

is released as heat. Calculate Δ H for a process in which a 5.8 g sample of methane is

burned at constant pressure.

1 mole 890 KJ

I have to get the new no. of moles that burned to get the amount of energy released

No. of mole = Mass / M.Wt

= 5.8 / (12+4)

= 0.3625 mole

1 mole 890 KJ

0.3625 mole ???? KJ

??? KJ = 0.3625∗890

1 = 322.625 KJ



6. Consider the following reaction:

CH4(g) + 2 O2(g) CO2 (g) + 2 H2O (L) ΔH= -891 KJ

Calculated the enthalpy change for each of the following cases:

a- 1.00 g methane is burned in excess oxygen.

b- 1.00*103 L methane gas at 740 torr and 250C is burned in excess oxygen.

a- I have to get the no. of moles that burned to get the amount of energy released

No. of mole = Mass / M.Wt

= 1 / (16)

= 0.0625

enthalpy change is = No. of mole * -891

= 0.0625 * -891 = -55.6875 KJ

b- I have to get the no. of moles that burned to get the amount of energy released

P V = N R T

( atm ) ( Lit ) = ( g mole ) (𝑎𝑡𝑚 .𝑙𝑖𝑡

𝑔 𝑚𝑜𝑙 . 𝑘 ) ( oK )

760 torr ( mmHg) 1 atm

740 torr ( mmHg) ??? atm

??? atm = ( 740 * 1 ) / (760 ) = 0.9736 atm

PV=NRT

( 0.9736 atm ) ( 1,000 L ) = ( ?? g mole ) ( 0.082𝑎𝑡𝑚 .𝑙𝑖𝑡

𝑔 𝑚𝑜𝑙 . 𝑘 )( 298 oK )

( ?? g mole ) = (973.6)/(298*0.082)

( ?? g mole ) = (973.6)/(24.436)

( ?? g mole ) = 39.84

enthalpy change is = No. of mole * -891

= 39.84 * -891 = - 35,499.98 KJ



7. The enthalpy of combustion of solid Carbon to form carbon dioxide is -393.7

KJ/mole carbons, and the enthalpy of combustion of carbon monoxide to form

carbon dioxide is -283.3 KJ/mole carbon monoxide. Use these data to calculate Δ H

for the reaction:

2C (s) + O2 2 CO (g)

C (s) + 1

2O2 (g) CO (g) Δ H1 =????? KJ/mole { Find }

CO(s) + 1

2O2(g) CO2 (g) Δ H2 = -283.3 KJ/mole { Given }

_______________________________________ ________________________________________

C (s) + O2(g) CO2 (g) Δ H3 = -393.7 KJ/mole { Given }

Δ H3 = Δ H1 + Δ H2

-393.7 = Δ H1 + ( -283.3 )

Δ H1 = -393.7 + 283.3

Δ H1 = -110.4 KJ/mole

so that

C (s) + 1

2O2(g) CO (g) Δ H1 = -110.4 KJ/mole { I proved }

But I need

2C (s) + O2 2 CO (g) Δ H = 2* -110.4 KJ/mole

Δ H = - 220.8 KJ /mole



8. Ethanol (C2H5OH) has been proposed as an alternative fuel. Calculate the standard

of enthalpy of combustion per gram of liquid ethanol. Knowing that the standard

enthalpy of formation of ethanol (liquid) = -278 KJ/mole.

standard of enthalpy of combustion per gram of liquid ethanol

C2H5OH + 3 O2 2 CO2 + 3 H2O


Δ HC C2H5OH = (2 * Δ HF CO2 ) + (3 * Δ HF H20 ) + (-1 * Δ HF C2H5OH) + (- 3* Δ HF O2 )

Δ HC C2H5OH = (2 * -394 ) + (3 * -286 ) + (-1 * -278 ) + (- 3 * 0 )

Δ HC C2H5OH = (- 788 ) + (- 858 ) + ( 278 ) + (0 ) = - 1,368 KJ/mole

Mass C2H5OH = NO. Mole C2H5OH * M.WT C2H5OH

Mass C2H5OH = 1 * ( 2*12+1*6+1*16) = 46 gm


= Δ HC C2H5OH

M.wt of methanol =

− 1,368

46 = - 29.739 KJ/gm



9. Some automobiles and buses have been equipped to burn propane (C3H8). Compare

the amounts of energy that can be obtained per gram of (C3H8 (g)) and per gram of

gasoline, assuming that gasoline is pure octane, (C8H18 (L)). Look up the boiling

point of propane. What disadvantages are there to using propane instead of

gasoline as fuel? (Standard enthalpies of formation of C3H8(g) and C8H18(L) are -104

and -269 KJ/mole respectively )

standard of enthalpy of combustion per gram of propane

C3H8 + 5 O2 3CO2 + 4 H2O


Δ HC C3H8 = (3 * Δ HF CO2 ) + (4 * Δ HF H20 ) + (-1 * Δ HF C3H8) + (-5 * Δ HF O2 )

Δ HC C3H8 = (3 * -394 ) + (4 * -286 ) + (-1 * -104 ) + (-5 * 0 )

Δ HC C3H8 = (-1,182 ) + ( -1,144 ) + ( 104 ) + ( 0 ) = -2,222 KJ/mole

Mass C3H8= NO. Mole C3H8* M.WT C3H8

Mass C3H8= 1 * (3*12+8) = 1*32 = 44 gm


= Δ HC CH 3OH

Mass CH 3OH =

− 2,222

44 = - 50.50 KJ/gm

standard of enthalpy of combustion per gram of octane

C8H18 + 25

2 O2 8 CO2 + 9 H2O



2 * Δ HF O2 )

Δ HC C8H18 = (8 * -394 ) + (9 * -286 ) + (-1 * -269 ) + (- 25

2 * 0 )

Δ HC C8H18 = ( - 3,152 ) + ( - 2,574 ) + ( 269 ) + ( 0 ) = - 5,457 KJ

Mass C8H18 = NO. Mole C8H18 * M.WT C8H18

Mass C8H18 = 1 * (8*12+1*18) = 114 gm

standard enthalpy of combustion per gram of Octane

= Δ HC C8H18

Mass C8H18 =

−5,457

114 = - 47.86 KJ/gm

Look up the boiling point of propane

Propane Boiling Point is -44𝑜F = - 6.66𝑜𝑐 = −44𝑜𝑓+32

1.8

What disadvantages are there to using propane instead of gasoline as fuel?

1- At the ambient temperature propane found in gas state, which is hard to handling.



10. Assume that -4.19*106 KJ of energy is needed to heat a home. If this energy is

derived from the combustion of methane (CH4), what volume of methane,

measured at STP must be burned? (Δ𝐻𝑜combustion for CH4= -891 KJ/mole)

1 mole - 891 KJ

??? Mole - 4.19*106 KJ

??? mole = (− 4.19∗106)∗1

− 891 = 4,702.581 mole

P V = N R T

(atm ) ( Lit ) = ( g mole ) (𝑎𝑡𝑚 .𝑙𝑖𝑡

𝑔 𝑚𝑜𝑙 . 𝑘 )( oK )

(1 atm ) ( ?? Lit) = (4,702.581 g mole) (0.082 𝑎𝑡𝑚 .𝑙𝑖𝑡

𝑔 𝑚𝑜𝑙 . 𝑘 ) (273 oK)

(?? Lit ) = 4,702.581∗0.082∗273

1 = 105,271.97 lit



11. At 298 K, the standard enthalpies of formation for C2H2 (g) and C6H6 (L) are 227

KJ/mole and 49 KJ/mole, respectively.

a- Calculate ΔHo for C6H6 (L) 3 C2H2 (g)

b- Both acetylene (C2H2) and benzene (C6H6) can be used as fuel. Which compound

will liberate more energy per gram when combusted in air?

a- Calculate ΔHo for C6H6 (L) 3 C2H2 (g)

standard enthalpies of formation for C2H2 (g)

2 C (S) + H2 (g) C2H2 (g) ΔHF C2H2= 227 KJ/mole

6 C (S) + 3H2 (g) 3C2H2 (g) ΔH F 3C2H2= 3*227 = 681 KJ

standard enthalpies of formation for C6H6 (L)

6 C (S) + 3H2 (g) C6H6 (L) ΔHF C6H6 = 49 KJ/mole

C6H6 (L) 6 C (S) + 3H2 (g) ΔHF C6H6 = - 49 KJ/mole

6 C (S) + 3H2 (g) 3C2H2 (g) ΔHF 3C2H2= 3*227 = 681 KJ

C6H6 (L) 6 C (S) + 3H2 (g) ΔHF C6H6 = 1*- 49 = -49 KJ

__________________________________________ ___________________________________

Δ𝐻𝑜 for C6H6 (L) 3 C2H2 (g) Δ𝐻𝑜 = 681 -49 = 632 KJ

b- Both acetylene (C2H2) and benzene (C6H6) can be used as fuel. Which compound will

liberate more energy per gram when combusted in air?

Combustion for C6H6 (L)

C6H6 (L) + 15

2O2 6 CO2 + 3H2O (g)


2 * Δ HF O2 )

Δ HC C6H6 = (6 * -394 ) + (3 * -286 ) + (-1 * 49) + (- 25

2 * 0 )

Δ HC C6H6 = ( - 2,364 ) + ( - 858 ) + ( -49) + ( 0 ) = - 3,271 KJ/mole


Mass C6H6= 1 * (6*12+6) = 1*78 = 78 gm


= Δ HC C6H6

Mass C6H6 =

− 3,271

78 = - 41.93 KJ/gm

Combustion for C2H2 (g)

C2H2 (g) + 5

2O2 2 CO2 + H2O (g)


2 * Δ HF O2 )



Δ HC C6H6 = (2 * -394 ) + (1 * -286 ) + (-1 * 227) + (- 5

2 * 0 )

Δ HC C6H6 = ( - 788 ) + ( - 286 ) + ( -227) + ( 0 ) = - 1,301 KJ/mole


Mass C2H2= 1 * (2*12+2) = 1*26 = 26 gm


= Δ HC C2H2

M.wt C2H2 =

− 1,301

26 = - 50.04 KJ/gm

Which compound will liberate more energy per gram when combusted in air?

(C2H2)



12. Combustion of table sugar produces CO2 (g) and H2O (L).

a- Assuming that sugar is pure sucrose, C12H22O11(S), write the balanced equation for

the combustion reaction.

b- Calculated ΔH in KJ/mole C12H22O11(S) for the combustion reaction of sucrose.

( Standard enthalpy of formation of C12H22O11(S) is 2244 KJ/mole)

a- Write the balanced equation for the Combustion reaction of C12H22O11(S)

C12H22O11(S) + 12 O2 12 CO2 + 11 H2O (g)

b- Calculated ΔH

C12H22O11(S) + 12 O2 12 CO2 + 11 H2O (g)

Δ HC C12H22O11(S) = ( 12 * Δ HF CO2 ) + ( 11 * Δ HF H20 ) + (-1 * Δ HF C2H2 ) + (- 12 * Δ HF O2 )

Δ HC C12H22O11(S) = ( 12 * -394 ) + ( 11 * -286 ) + (-1 * 2244) + (- 12 * 0 )

Δ HC C12H22O11(S) = ( - 4,728 ) + ( - 3,146 ) + ( -2244) + ( 0 ) = - 10,118 KJ/mole

thermo-chemistry enthalpy changes in chemical process part 1

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