The enthalpy change ∆ H for the reaction N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g ) is - 92 . 38 kJ at 298 K . What is ∆ U at 298 K ?

Given, N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g ) . ∆ H = - 92 .38 kJ . T = 298 K . ∆ U = ? Formula used in the calculation is as follows, ∆ H = ∆ U + ∆ n g RT . ∆ U = ∆ H - ∆ n g RT . Where, ∆ H = Enthalpy change ∆ U = Internal energy change ∆ n g = Change in number of gas species (Number of moles). ∆ n g = 2 - 4 = - 2 . R = Gas Constant 8 . 314 J K - 1 mol - 1 . T = Temperature. So, ∆ U = ∆ H - ∆ n g RT . ∆ U = - 92 .38 - - 2 8 .314 × 0 .001 × 298 . ∆ U = - 92 . 38 + 4 . 955 = - 87 . 425 kJ . [ ∵ Multiplied by 0 . 001 to Change joule into kJ ].

The internal energy change ∆ U for the reaction CH 4 ( g ) + 2 O 2 g → CO 2 ( g ) + 2 H 2 O ( l ) is – 885 kJ mol - 1 at 298 K . What is ∆ H at 298 K ?

Given, CH 4 ( g ) + 2 O 2 ( g ) → CO 2 ( g ) + 2 H 2 O l . ∆ U = - 885 kJ mol - 1 . T = 298 K . ∆ H = ? Formula used for the calculation is ∆ H = ∆ U + ∆ n g RT . Where, ∆ H = Enthalpy ∆ U = Internal Energy. ∆ n g = Change in number of gas species (Number of moles). ∆ n g = 1 - 3 = - 2 . R = Gas Constant ( 8 . 314 J K - 1 mol - 1 ). T = Temperature Change in enthalpy: ∆ H = ∆ U + ∆ n g RT . ∆ H = ( - 885 ) + ( - 2 ) 8 .314 × 0 .001 × 298 ∵ 1 J = 0 . 001 kJ . ∆ H = - 885 - 4 . 955 = - 889 . 955 kJ .

E M B I B E

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 1

HARD

11th CBSE

IMPORTANT

Earn 100

The heat of combustion of ${CH}_{4}$ $(g)$ at constant volume is measured in a bomb calorimeter at $298.2 K$ and found to be $- 885389 J mol^{- 1}$ . Find the value of enthalpy change.

Important Questions on Thermodynamics

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 2

The enthalpy change $(∆ H)$ for the reaction $N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g)$ is $- 92.38 kJ$ at $298 K$ . What is $∆ U$ at $298 K$ ?

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 3

The internal energy change

(∆ U)

for the reaction

{CH}_{4} (g) + 2 O_{2} (g) \to {CO}_{2} (g) + 2 H_{2} O (l)

is

- 885 kJ {mol}^{- 1}

at

298 K

. What is

∆ H

at

298 K

?

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 4

When

{NH}_{4} {NO}_{2}

(s)

decomposes at

373 K

, it forms

N_{2} (g)

and

H_{2} O (g)

. The

∆ H

for the reaction at one atmospheric pressure and

373 K

is

- 223.6 kJ {mol}^{- 1}

of

{NH}_{4} {NO}_{2}

(s)

decomposed. What is the value of

∆ U

for the reaction under the same conditions? Given,

(R = 8.31 J K^{- 1} {mol}^{- 1})

.

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 5

When

0.532 g

of benzene

(C_{6} H_{6})

, boiling point

353 K

, is burnt with excess of oxygen in a constant volume system,

22.3 kJ

of heat is given out. Calculate

∆ H

for the combustion process.

(R = 8.31 J K^{- 1} {mol}^{- 1})

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 6

The heat of Combustion of naphthalene

(C_{10} H_{8} (s))

at constant volume was found to be

- 5133 kJ {mol}^{- 1}

. Calculate the value of enthalpy change.

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 7

The molar heat of formation of

{NH}_{4} {NO}_{2} (s)

is

- 367.54 kJ

and those of

N_{2} O (g)

and

H_{2} O (l)

are

+ 81.46 kJ

and

- 285.78 kJ

respectively at

25 ° C

and

1.0

atmospheric pressure. Calculate the

ΔH

and

ΔU

for the reaction.

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 8

Determine the value of $ΔH$ and $ΔU$ for the reversible isothermal evaporation of $90.0 g$ of water at $100 ° C$ . Assume that water vapour behave as an ideal gas and heat of evaporation of water is $500$ $Cal g^{- 1}$ $(R = 2.0 Cal {mol}^{- 1} K^{- 1})$ .