﻿ The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 250C; the heat of combustion (in kJ mol–1) of benzene at constant pressure will be (R = 8.314 JK–1 mol–1) | Thermodynamics

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# 101.The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 250C; the heat of combustion (in kJ mol–1) of benzene at constant pressure will be (R = 8.314 JK–1 mol–1)–3267.6 4152.6 –452.46 3260

A.

–3267.6

Δng = 6 - 7.5
= -1.5 (change in gaseous mole)
ΔU or ΔE = - 3263.9 kJ
ΔH = ΔU + ΔngRT
Δng = - 1.5
R = 8.314 JK-1 mol-1
T = 298 K
So ΔH = -3263.9 + (-1.5) 8.314 x 10-3 x 298
= -3267.6 kJ
ΔH = Heat at constant pressure
ΔU/ΔE = Heat at constant volume
R = gas constant

102.

Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction?

• A and D

• A and B

• B and C

• C and D

B.

A and B

Therefore ln K vs 1/T the graph will be a straight line with slope equal to $-\frac{∆{H}^{°}}{R}$.Since reaction is
exothermic, therefore $∆{H}^{°}$ itself will be negative resulting in positive slope.

103.

Identify the invalid equation

C.

Relation between heat of reaction ($∆$rH° ) and bond enthalpies of reactants and products is

104.

For the process A (1, 0.05 atm, 32°C) → A (g, 0.05 atm, 32°C)

The correct set of thermodynamic parameters is

• $∆$G = 0 and $∆$S = -ve

• $∆$G = 0 and $∆$S = +ve

• $∆$G = +ve and $∆$S = 0

• $∆$G = -ve and $∆$S = 0

B.

$∆$G = 0 and $∆$S = +ve

(i) Since, n and T are constants.

G = 0

Hence, the system is equilibrium.

(ii) Since, P → decreases

S → increases, i.e. it is +ve

Therefore, among all the options given, the correct answer is in option b, i.e. $∆$G = 0 and $∆$S = +ve.

105.

Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

• (a) 195 K (b) 2.7 kJ

• (a) 189 K (b) 2.7 kJ

• (a) 195 K (b) –2.7 kJ

• (a) 189 K (b) – 2.7 kJ

D.

(a) 189 K (b) – 2.7 kJ

106.

The first law of thermodynamics for isothermal process is

• q = -W

• $∆$U = W

• $∆$U = qv

• $∆$U = -qv

A.

q = -W

According to first law of thermodynamics,

$∆$U = q + W

where, $∆$U = internal energy; q = Heat and W = work done

For isothermal process,

$∆$T = 0; $∆$U = 0

Therefore, q = -W