﻿ Engineering Entrance Exam Question and Answers | Thermodynamics - Zigya

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# Thermodynamics

#### Multiple Choice Questions

91.

Δ U is equal to

• Isochoric work

• Isobaric work

C.

From 1st law thermodynamics:
ΔU = q + w
q = 0
∴ ΔU = w
∴ Work involves in the adiabatic process is at the expense of a change in internal energy of the system.

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92.

The condition for a reaction to occur spontaneously is

• $∆$H must be negative

• $∆$S must be negative

• $\left(∆\mathrm{H}-\mathrm{T}∆\mathrm{S}\right)$ must be negative

• $\left(∆\mathrm{H}+\mathrm{T}∆\mathrm{S}\right)$ must be negative

C.

$\left(∆\mathrm{H}-\mathrm{T}∆\mathrm{S}\right)$ must be negative

93.

The heats of combustion of carbon and carbon monoxide are −393.5 and −283.5 kJ mol−1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is:

• 676.5

• -676.5

• -110.5

• -110.5

C.

-110.5

C(s) + O2 (g) → CO2 (g); ΔH = -393.5 kJ mol-1 ... (i)

CO + O2/2 → CO2 (g); ΔH = - 283.5 kJ mol-1 ....(ii)
On subtracting Eq. (ii) from Eq. (i), we get

C (s) + O2/2 (g) → CO (g)

ΔH = (-393.5 + 283.5) kJ mol-1 = - -110 kJ mol-1

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94.

In which of the following reactions will $∆$U be equal to $∆$H?

B.

95.

Thermal decomposition of ammonium dichromate gives

• N2, H2O and Cr2O3

• N2, NH3 and CrO

• (NH4)2CrO4 and H2O

• N2, H2O and CrO3

A.

N2, H2O and Cr2O3

Thermal decomposition of ammonium dichromate (NH4)2Cr2O7 on heating gives N2, Cr2O3 and H2O as follows:

(NH4)2Cr2O$\stackrel{∆}{\to }$N2 + Cr2O3 + 4H2O

96.

Which of the following plots represent an exothermic reaction?

A.

For exothermic reaction,

97.

Given
C(grahite) + O2(g) → CO2(g)
ΔrH°  = - 393.5 kJ mol-1
H2(g) + 1/2O2(g) → H2O (l)
ΔrH°  = +890.3 kJ mol-1
Based on the above thermochemical equations, the value of ΔrH° at 298 K for the reaction
C(grahite) + 2H2(g) →CH4 will be

•  +74.8 kJ mol–1

• +144.0 kJ mol–1

• –74.8 kJ mol–1

• –74.8 kJ mol–1

C.

–74.8 kJ mol–1

CO2(g) + 2H2O (l)→ CH4 (g) + 2O2(g);
ΔrH°= 890.3
ΔfH° –393.5 –285.8 ? 0
ΔrH°= Σ(ΔfH°)products -Σ(ΔfH°)reactant
890.3 = [ 1 x(ΔfH°)CH4 + 2x0]-[1x(-393.5)+2(-285.8)]
fH°)CH4 = 890.3-965.1 = -74.8 kJ/mol

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98.

For the reaction

N(g) + 3H2 (g) $⇌$ 2NH3 (g) $∆$H is equal to

• $∆$E - 2RT

• $∆$E - RT

• $∆$E + RT

• $∆$E + 2RT

A.

$∆$E - 2RT

For the reaction,

N2 (g) + 3H2 (g) $⇌$ 2NH3 (g)

($\because$ $∆$H = $∆$E + $∆$ng . RT)

$∆$n = 2 - 4 = -2

$\therefore$

99.

The enthalpy of combustion of glucose (mol.wt.: 180 g mol-1) is -2840 kJ mol-1. Then the amount of heat evolved when 0.9 g ofglucose is burnt, will be

• 14.2 kJ

• 142 kJ

• 28.4 kJ

• 1420 kJ

A.

14.2 kJ

The enthalpy of combustion of glucose (mol.wt. = 180 g/mol) = -2840 kJ/ mol

180 g glucose gives enthalpy of combustion = 2840 kJ/ mol

1 gm glucose gives enthalpy of combustion = $\frac{2840}{180}$

So, 0.9 gm glucose gives enthalpy of combustion

$\frac{2840}{200}\mathrm{kJ}$

= 14.2 kJ

Therefore, -14.2 kJ is the amount of heat evolved when 0.9 gm of glucose is burnt.

100.

The work done during combustion of 9 × 10-2 kg of ethane, C2H6 (g) at 300 K is (Given R = 8.314 J deg-1 mol-1, atomic mass C = 12, H= 1).

• 6.236 kJ

• -6.236 kJ

• 18.71 kJ

• -18.71 kJ

C.

18.71 kJ

Work done in a chemical reaction,

W = $∆$ngRT

where, $∆$ng = number of moles of gaseous products - number of moles of gaseous reactants.

Combustion of ethane is-

C2H6 (g) + $\frac{7}{2}$O2 (g) → 2CO2 (g) + 3H2O (l)

$\therefore$ng = 2 - 4.5 = -2.5

W = (2.5 mol) (8.314 JK-1 mol-1) × 300 K

= 6235.5 kJ = 6.2355 J

Now, 1 mole of C2H6 = 30 gm of C2H6 = 6.2355 kJ

Work done for combustion of 30 gm of C2H= 6.2355 kJ

$\therefore$ Work done for combustion of 90 gm of C2H

= 18.7065 kJ = 18.71 kJ