Calculate the standard enthalpy of formation of C 2 H 6 from the following data: 2 C 2 H 6 ( g ) + 7 O 2 ( g ) → 4 CO 2 ( g ) + 6 H 2 O ( 1 ) , ∆ H ° = - 3119 kJ ∆ f H ° ( CO 2 ) = - 393 . 5 kJ mol - 1 , ∆ f H ° ( H 2 O ) = - 285 . 8 kJ mol - 1 .

2 C 2 H 6 ( g ) + 7 O 2 ( g ) → 4 CO 2 ( g ) + 6 H 2 O ( 1 ) , ∆ H ° = - 3119 kJ given, ∆ H f ° ( CO 2 ) = - 393 . 5 KJ / mole and ∆ H f ° ( H 2 O ) = - 285 . 8 KJ / mole we know that, ∆ H ° = ΣH f ° ( product ) - ΣH f ° ( Reactants ) ∆ H ° = 4 ∆ H f ° ( CO 2 ) + 6 ∆ H f ° ( H 2 O ) - 2 ∆ H f ° ( C 2 H 6 ) + 7 ∆ H f ° ( O 2 ) - 3119 = 4 ( - 393 . 5 ) + 6 ( - 285 . 8 ) - 2 ∆ H f ° ( C 2 H 6 ) + 0 ( ∆ H f ° ( O 2 ( g ) ) = 0 ) 2 ∆ H f ° ( C 2 H 6 ) = + 3119 - [ 3288 . 8 ] 2 ∆ H f ° ( C 2 H 6 ) = - 169 . 8 ∆ H f ° ( C 2 H 6 ) = - 82 . 9 KJ / mole

Calculate the standard enthalpy of formation of C 2 H 6 from the following data: 2 C 2 H 6 ( g ) + 7 O 2 ( g ) → 4 CO 2 ( g ) + 6 H 2 O ( 1 ) , ∆ H ° = - 3119 kJ ∆ f H ° ( CO 2 ) = - 393 . 5 kJ mol - 1 , ∆ f H ° ( H 2 O ) = - 285 . 8 kJ mol - 1 .

2 C 2 H 6 ( g ) + 7 O 2 ( g ) → 4 CO 2 ( g ) + 6 H 2 O ( 1 ) , ∆ H ° = - 3119 kJ given, ∆ H f ° ( CO 2 ) = - 393 . 5 KJ / mole and ∆ H f ° ( H 2 O ) = - 285 . 8 KJ / mole we know that, ∆ H ° = ΣH f ° ( product ) - ΣH f ° ( Reactants ) ∆ H ° = 4 ∆ H f ° ( CO 2 ) + 6 ∆ H f ° ( H 2 O ) - 2 ∆ H f ° ( C 2 H 6 ) + 7 ∆ H f ° ( O 2 ) - 3119 = 4 ( - 393 . 5 ) + 6 ( - 285 . 8 ) - 2 ∆ H f ° ( C 2 H 6 ) + 0 ( ∆ H f ° ( O 2 ( g ) ) = 0 ) 2 ∆ H f ° ( C 2 H 6 ) = + 3119 - [ 3288 . 8 ] 2 ∆ H f ° ( C 2 H 6 ) = - 169 . 8 ∆ H f ° ( C 2 H 6 ) = - 82 . 9 KJ / mole

HARD

Earn 100

Calculate the standard enthalpy of formation of $C_{2} H_{6}$ from the following data:

$2 C_{2} H_{6} (g) + 7 O_{2} (g) \to 4 {CO}_{2} (g) + 6 H_{2} O (1), ∆ H ° = - 3119 kJ$

$∆_{f} H ° ({CO}_{2}) = - 393.5 kJ {mol}^{- 1},$

$∆_{f} H ° (H_{2} O) = - 285.8 kJ {mol}^{- 1} .$

Important Questions on Thermodynamics

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

When $0.8 g$ of glucose, $C_{6} H_{12} O_{6}$ , was burnt in a bomb calorimeter, according to the following equation, the temperature raise is found to be $2 K$ at $1 atm$ .

$C_{6} H_{12} O_{6 (s)} + 6 O_{2 (g)} ⟶ 6 {CO}_{2 (g)} + 6 H_{2} O_{(l)}$

If the heat capacity of the bomb calorimeter is $6.8 kJ K^{- 1}$ , the approximate enthalpy change of the reaction in $kJ {mol}^{- 1}$ is (Molar mass of glucose $= 180 g {mol}^{- 1}$ )

HARD

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in $K$ ) at which the reduction of cassiterite by coke would take place.

$298 K : Δ_{f} H^{0} ({SnO}_{2} (s)) = - 581.0 kJ {mol}^{- 1}, Δ_{f} H^{0} ({CO}_{2} (g)) = - 394.0 kJ {mol}^{- 1}$

$S^{0} ({SnO}_{2} (s)) = 56.0 {JK}^{- 1} {mol}^{- 1}, S^{0} (Sn (s)) = 52.0 {JK}^{- 1} {mol}^{- 1}$

$S^{0} (C (s)) = 6.0 {JK}^{- 1} {mol}^{- 1}, S^{0} ({CO}_{2} (g)) = 210.0 {JK}^{- 1} {mol}^{- 1}$

Assume that the enthalpies and the entropies are temperature independent.

EASY

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

The heat of reaction for

$C_{10} H_{8} (s) + 12 O_{2} (g) ⟶ 10 {CO}_{2} (g) + 4 H_{2} O (l)$ at constant volume is $- 1228.2 kcal$ at $25^{0} C$ . The heat of reaction at constant pressure and same temperature is

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

The standard heat of formation of

{CH}_{4}, {CO}_{2}

and

H_{2} O

are

- 762, - 394.8

and

- 285.82 kJ {mol}^{- 1},

respectively. Heat of vaporisation of water is

44 kJ {mol}^{- 1} .

Calculate the amount of heat eveolved when

22.4 L

{CH}_{4},

kept under normal conditions, is oxidized into its gaseous products.

EASY

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

For a reaction,

A (g) \to A (l); ΔH = - 3 RT .

The correct statement for the reaction is

EASY

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

Standard molar enthalpies of formation of

{CaCo}_{3} (s), Ca ω (s);

and

{Co}_{2} (g)

are

- 1206.92 KJ {mol}^{- 1}, - 635.09 KJ {mol}^{- 1}

and

- 393.51

respectively. The

Δ

Hr for decomposition of

Ca {CO}_{3} (s)

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

For the reaction

$N_{2} O_{4} (g) ⟶ 2 {NO}_{2} (g)$

EASY

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

For which of the following reactions,

∆ H

is equal to

∆ U

HARD

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

For complete combustion of ethanol,

C_{2} H_{5} OH (l) + 3 O_{2} (g) \to 2 {CO}_{2} (g) + 3 H_{2} O (l)

,
the amount of heat produced as measured in bomb calorimeter, is

{1364.47 kJ mol}^{-1}

25 ℃

. Assuming ideality the Enthalpy of combustion,

Δ_{c} H

, for the reaction will be:

(R = {8.314 kJ mol}^{-1})

EASY

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

The difference between

∆ H

and

∆ U

(∆ H - ∆ U)

, when the combustion of one mole of heptane

(l)

is carried out at a temperature

T

, is equal to:

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

If the standard molar enthalpy change for combustion of graphite powder is

- 2.48 \times 10^{2} kJ {mol}^{- 1}

, the amount of heat generated on combustion of

1 g

of graphite powder in

kJ

(Nearest integer):

HARD

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

Given:

C (graphite) + O_{2} (g) \to {CO}_{2} (g); Δ_{r} H^{o} = - 393 .5 kJ {mol}^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ_{r} H^{o} = - 285 .8 kJ {mol}^{- 1}

{CO}_{2} (g) + 2 H_{2} O (l) \to {CH}_{4} (g) + 2 O_{2} (g); Δ_{r} H^{o} = + 890 .3 kJ {mol}^{- 1}

Based on the above thermochemical equations, the value of

Δ_{r} H^{o}

298 K

for the reaction

C (graphite) + 2 H_{2} (g) \to {CH}_{4} (g)

will be:

EASY

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

17.0 g

{NH}_{3}

completely vapourises at

- 33.42 ° C

and

1

bar pressure and the enthalpy change in the process is

23.4 kJ {mol}^{- 1}

. The enthalpy change for the vapourisation of

85 g

{NH}_{3}

under the same conditions is

kJ

HARD

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

Nitroglycerine $(MW = 227.1)$ detonates according to the following equations,

$2 C_{3} H_{5} {({NO}_{3})}_{3} (l) ⟶ 3 N_{2} (g) + \frac{1}{2} O_{2} (g) + 6 {CO}_{2} (g) + 5 H_{2} O (g)$

The standard molar enthalpies of formation, ${ΔH}_{f} °$ for all the compounds are given below.

${ΔH}_{f} ° [C_{3} H_{5} {({NO}_{3})}_{3}] = - 364 kJ / mol$
${ΔH}_{f} ° [{CO}_{2} (g)] = - 393.5 kJ / mol$
${ΔH}_{f} ° [H_{2} O (g)] = - 241.8 kJ / mol$
${ΔH}_{f} ° [N_{2} (g)] = 0 kJ / mol$
${ΔH}_{f} ° [O_{2} (g)] = 0 kJ / mol$

The enthalpy change when $10 g$ of nitroglycerine is detonated is

HARD

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

The bond dissociation energies of

X_{2} Y_{2}

and

XY

are in the ratio of

1 : 0 .5 : 1 . ΔH

for the formation of

XY

- 200 kJ {mol}^{- 1}

. What will be the bond dissociation energy of

X_{2}

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

Enthalpy of sublimation of iodine is

24 c a l g^{- 1}

200 ° C

. If specific heat of

I_{2}

(s) and

I_{2}

(vap) are

0.055

and

0.031 c a l g^{- 1} K^{- 1}

respectively, then enthalpy of sublimation of iodine at

250 ° C

c a l g^{- 1}

is:

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

The heats of combustion of carbon and carbon monoxide are

- 393.5 a n d - 283.5 k J m o l^{- 1},

respectively. The heat of formation

(in kJ)

of carbon monoxide per mole is:

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

For the reaction:

X_{2} {O_{4}}_{(l)} \to 2 X {O_{2}}_{(g)}

∆ U = 2 .1 kcal, ∆ S = 20 cal K^{- 1}

300 K

Hence,

∆ G

is:

MEDIUM

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

One mole of ethanol is produced reacting graphite,

H_{2}

and

O_{2}

together. The standard enthalpy of formation is

- 277.7 k J m o l^{- 1} .

Calculate the standard enthalpy of the reaction when

4

moles of graphite is involved

EASY

Chemistry>Physical Chemistry>Thermodynamics>Enthalpy Change of a Reaction- Reaction Enthalpy

The enthalpy change on freezing of

1 mol

of water at

5 ° C

to ice at

- 5 ° C

is:

(Given

Δ_{fus} H = 6 kJ {mol}^{- 1}

0 ° C,

C_{p} (H_{2} O, l) = 75 .3 J {mol}^{- 1} K^{- 1}

C_{p} (H_{2} O, s) = 36 .8 J {mol}^{- 1} K^{- 1}

)

Calculate the standard enthalpy of formation of C2H6 from the following data: 2C2H6 (g) + 7 O2 (g) → 4CO2 (g)+6 H2O (1), ∆H° = -3119kJ ∆fH° (CO2) = -393.5 kJ mol-1, ∆fH° (H2O) = -285.8 kJ mol-1.

Important Questions on Thermodynamics

Learn and Prepare for any exam you want !

Calculate the standard enthalpy of formation of $C_{2} H_{6}$ from the following data:

$2 C_{2} H_{6} (g) + 7 O_{2} (g) \to 4 {CO}_{2} (g) + 6 H_{2} O (1), ∆ H ° = - 3119 kJ$

$∆_{f} H ° ({CO}_{2}) = - 393.5 kJ {mol}^{- 1},$

$∆_{f} H ° (H_{2} O) = - 285.8 kJ {mol}^{- 1} .$