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# Evaluate Yourself: Thermodynamics(Chemistry)

Chemistry : Thermodynamics : Evaluate Yourself

Evaluate Yourself

1. Calculate ∆Hf° for the reaction

CO2(g)+ H2(g) → CO(g)+ H2O(g)

given that ∆Hf° for CO2 (g), CO (g) and H2O(g) are – 393.5, – 111.31 and – 242 kJ mol–1 respectively.

Solution :

Given

∆Hf0 CO2 = -393.5 kJ mol-1

∆Hf0 CO = -111.31 kJ mol-1

∆Hf0 (H2O) = -242 kJ mol-1

CO2(g) + H2(g) CO(g) + H2O(g)

∆Hr0 = ?

∆Hr0 = Σ (∆Hf0 )products– Σ (∆Hf0)reactants

∆Hr0 = [∆Hf0 (CO) + ∆Hf0 (H2O)] – [∆Hf0 (CO2 )+∆Hf0

(H2)] ∆Hr0 = [– 111.31 + (– 242)] – [– 393.5 + (0)]

∆Hr0 = [– 353.31] + 393.5

∆Hr0 = 40.19

∆Hr0 = + 40.19 kJ mol–1

2. Calculate the amount of heat necessary to raise 180 g of water from 25° C to 100° C. Molar heat capacity of water is 75.3 J mol-1 K-1

Solution :

Given :

number of moles of water n

=180g / (18g mol-1  )= 10 mol

molar heat capacity of water

CP = 75.3 J K–1 mol–1

T2 = 1000 C = 373 K

T1 = 250 C= 298 K

ΔH = nCP (T2 – T1)

ΔH = 10 mol × 75.3 J mol–1 K–1 × (373 – 298) K

ΔH = 56475 J

ΔH = 56.475 kJ

3. From the following data at constant volume for combustion of benzene, calculate the heat of this reaction at constant pressure condition.

C6H6(l) + 7½ O2(g) → 6CO2 (g) + 3 H2O(l)

ΔU at 25° C = -3268.12 kJ

Solution :

T = 250 C = 298 K ;

∆U = –3268.12 kJ mol–1

∆H = ?

∆H = ∆U + ∆ngRT

∆H = ∆U + (np – nr) RT

H = -3268.12+ (6 - 7/2) x 8 314 × 10  -3 x 298.

∆U = –3268.12  + (2.5 × 8.314 ×10-3 × 298)

∆U = – 3268.12 + 6.19

∆U = – 3261.93 kJ mol–1

4. When a mole of magnesium bromide is prepared from 1 mole of magnesium

and 1 mole of liquid bromine, 524 kJ of energy is released.

The heat of sublimation of Mg metal is 148 kJ mol–1. The heat of dissociation of bromine gas into atoms is 193 kJ mol–1. The heat of vapourisation of liquid bromine is 31 kJ mol–1. The ionisation energy of magnesium is 2187 kJ mol–1 and the electron affinity of bromine is – 662 kJ mol–1. Calculate the lattice energy of magnesium bromide.

Solution :

Given :

Mg(S) + Br2(1)

MgBr2(S) Hf0 = –524 KJ mol–1

Sublimation :

Mg(S) Mg(g)

H10 = +148 KJ mol–1

Ionisation:

Mg(g) Mg2+(g) + 2e

H20 = 2187 KJ mol–1

Br2(1) Br2(g)

H30 = +31 KJ mol–1

Dissociation :

Br2(g) 2Br(g)

H40 = +193 KJ mol–1

Electron affinity:

Br(g) + e Br (g)

H50 = – 331 KJ mol–1

Solution : ΔHf  =ΔH1 + ΔH2 + ΔH3 + ΔH4 +2ΔH5 + u

– 524 =   148 + 2187 + 31 + 193+(2 × – 331) + u

– 524 =   1897 + u

u =–524 – 1897

u =– 2421 kJ mol–1

5. An engine operating between 1270 C and 470 C takes some specified amount of heat from a high temperature reservoir. Assuming that there are no frictional losses, calculate the percentage efficiency of an engine.

Solution:

Given

Th = 1270 C = 127 + 273 = 400

K TC = 470 C = 47 + 273 = 320

K % efficiency η = ? 6. Urea on hydrolysis produces ammonia and carbon dioxide. The standard entropies of urea, H2O, CO2, NH3 are 173.8, 70, 213.5 and 192.5J mole-1K-1 respectively. Calculate the entropy change for this reaction.

Solution: 7. Calculate the entropy change when 1 mole of ethanol is evaporated at 351 K. The molar heat of vaporisation of ethanol is 39.84 kJ mol-1

Solution: 8. For a chemical reaction the values of ΔH and ΔS at 300K are − 10 kJ mole−1 and −20 J deg−1 mole−1 respectively. What is the value of ΔG of the reaction? Calculate the ΔG of a reaction at 600 K assuming ΔH and ΔS values are constant. Predict the nature of the reaction.

Solution:

Given:

ΔH = –10 kJ mol–1 = –10000 J mol–1

ΔS = – 20 J K–1 mol–1

T =300 K

ΔG =?

ΔG =ΔH – TΔS

ΔG = – 10 kJ mol–1 – 300 K × (–20x10-3) kJ K–1 mol–1

ΔG =(–10+6) kJ mol–1

ΔG =– 4 kJ mol–1

At 600 K

ΔG =– 10 kJ mol–1 – 600 K

(–20x10-3) kJ K–1 mol–1

ΔG = (–10 + 12) kJ mol–1

ΔG = + 2 kJ mol–1

The value of ∆G is negative at 300K and the reaction is spontaneous, but at 600K the value ∆G becomes positive and the reaction is non spontaneous.

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