Thermodynamics — Page 2 | JEE Chemistry

Q11

For silver, Cp(J K–1 mol–1) = 23 +0.01 T. If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of

\Delta H

will be close to :

A 62 KJ

B 16 KJ

C 13 KJ

D 21 KJ

Correct Answer

Option A

Solution

Give that, n = 3 T1 = 300 T2 = 1000 Cp = 23 + 0.01T We know,

\Delta

H =

\int\limits_{{T_1}}^{{T_2}} {n{C_p}dT}

=

\int\limits_{300}^{1000} {3\left( {23 + {T \over {100}}} \right)dT}

=

3\left[ {23T + {{{T^2}} \over {200}}} \right]_{300}^{1000}

= 3[ 23(1000 - 300 +

{{1 \over {200}}}

((1000)2 - (300)2)] = 3[ 23 $\times$ 700 +

{{700 \times 1300} \over {200}}

] = 61950 J = 61.95 kJ

\simeq

62 kJ

Q12

The heat required to raise the temperature of body by 1 K is called :

A specific heat

B thermal capacity

C water equivalent

D none of these

Correct Answer

Option B

Solution

The heat required to raise the temperature of body by

1K

is called thermal capacity or heat capacity.

Q13

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant Kc is :

A -

\Delta G

= RT ln Kc

B

\Delta G^o

= RT ln Kc

C -

\Delta G^o

= RT ln Kc

D

\Delta G

= RT ln Kc

Correct Answer

Option C

Solution

\Delta {G^ \circ } = - RT\,1n{K_c}\,\,\,

or

\,\,\, - \Delta {G^ \circ } = RT\,1n{K_c}

Q14

At 320 K, a gas A2 is 20% dissociated to A(g). The standard free energy change at 320 K and 1 atm in J mol-1 is approximately : (R = 8.314 JK-1 mol-1; ln2 = 0.693; ln 3 = 1.098)

A 4763

B 2068

C 1844

D 4281

Correct Answer

Option D

Solution

A2 (g) $\rightleftharpoons$ 2 A (g) Assume initially concentration of A2 = [A2] = 1 m at equilibrium [A2] = 1 $\times$

{{80} \over {100}}

= 0.8 M and 20% of [A2] = 1 $\times$

{{20} \over {100}}

= 0.2 M

\therefore\,\,\,\,

[A] = 2 $\times$ 0.2 = 0.4 M Equilibrium constant K =

{{{{[A]}^2}} \over {[{A_2}]}}

=

{{{{\left[ {0.4} \right]}^2}} \over {\left[ {0.8} \right]}}

= 0.2

\Delta

Go = $-$ RT

\ell

nK = $-$ 8.314 $\times$ 320 $\times$

\ell

n(0.2) = 4281 J/mol.

Q15

For which of the following processes,

\Delta

S is negative ?

A H2(g)

\to

2H(g)

B N2(g, 1 atm)

\to

N2(g, 5 atm)

C C(diamond)

\to

C(graphite)

D N2(g, 273 K)

\to

N2(g, 300 K)

Correct Answer

Option B

Solution

N2(g, 1 atom)

\overset{\,}\longrightarrow

N2(g, 5 atom) Here pressure increases.

When pressure increases then the molecules of will come closer and intermoleculer distance decreases, so entropy will also decreases and

\Delta S\, < \,0

.

Q16

Lattice enthalpy and enthalpy of solution of NaCl are 788 kJ mol–1, and 4 kJ mol–1, respectively. The hydration enthalpy of NaCl is :

A –780 kJ mol–1

B –784 kJ mol–1

C 780 kJ mol–1

D 784 kJ mol–1

Correct Answer

Option B

Solution

\Delta

Hsol = Lattice enthalpy +

\Delta

Hhyd $\Rightarrow$ 4 = 788 +

\Delta

Hhyd $\Rightarrow$

\Delta

Hhyd = –784 kJ mol–1

Q17

Which of the following relations are correct? (A)

\mathrm{\Delta U=q+p\Delta V}

(B)

\mathrm{\Delta G=\Delta H-T\Delta S}

(C)

\Delta \mathrm{S}=\dfrac{q_{rev}}{T}

(D)

\mathrm{\Delta H=\Delta U-\Delta nRT}

Choose the most appropriate answer from the options given below :

A A and B only

B B and C only

C C and D only

D B and D only

Correct Answer

Option B

Solution

(A)

\mathrm{\Delta U=q-p\Delta V}

(B)

\mathrm{\Delta G=\Delta H-T\Delta S}

(C)

\mathrm{\Delta S=\frac{q_{rev}}{T}}

(D)

\mathrm{\Delta H=\Delta U+(\Delta nRT)}

Hence, (B) and (C) relations are correct.

Q18

If 100 mole of H2O2 decompose at 1 bar and 300 K, the work done (kJ) by one mole of O2(g) as it expands against 1 bar pressure is : 2H2O2(l)

\rightleftharpoons

2H2O(l) + O2(g) (R = 8.3 J K

-

1 mol

-

1)

A 62.25

B 124.50

C 249.00

D 498.00

Correct Answer

Option C

Solution

For reaction : 2H2O2(1) $\rightleftharpoons$ 2H2O(1) + O2(g) Given, p = 1 bar, T = 300 K, n = 100 mol, R = 8.3 J K $-$ mol $-$ 1 We know that w = p

\Delta

V = nRT = 100 $\times$ 8.3 $\times$ 300 = 249000 J or 249 kJ

Q19

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 :

A 676.5

B -676.5

C –110

D 110.5

Correct Answer

Option C

Solution

Given

C\left( s \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right);

\Delta H = - 393.5\,\,kJ\,mo{l^{ - 1}}......\left( i \right)

CO\left( g \right) + {1 \over 2}{O_2}\left( g \right) \to C{O_2}\left( g \right);

\Delta H = - 283.5\,kJ\,mo{l^{ - 1}}\,\,\,\,\,....\left( {ii} \right)

$\therefore$ Heat of formation of

Co = eqn\left( i \right) - eqn\left( {ii} \right)

= - 393.5 - \left( { - 283.5} \right)

= - 110\,kJ

Q20

If an endothermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point, then :

A

\Delta H

is -ve,

\Delta S

is +ve

B

\Delta H

and

\Delta S

are both +ve

C

\Delta H

and

\Delta S

are both -ve

D

\Delta H

is +ve,

\Delta S

is -ve

Correct Answer

Option B

Solution

TIPS/Formulae :

\Delta G = \Delta H - T\Delta S

Since

\Delta G = \Delta H - T\Delta S

for an endothermic reaction,

\Delta H = + ve

and at low temperature

\Delta S = + ve

Hence

\Delta G = \left( + \right)\Delta H - T\left( + \right)\Delta S

and if

T\Delta S < \Delta H

(at low temp)

\Delta G = + ve

(non spontaneous) But at high temperature, reaction becomes spontaneous i.e.

\Delta G = - ve.

because at higher temperature

T\Delta S > \Delta H.