Thermodynamics Test

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Thermodynamics Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero)

A
Cylic process $$:q=-w$$

B
Isothermal process $$:q=-w$$

C
Adiabatic process: $$\Delta U=-w$$

D
Isochoric process: $$\Delta U=q$$

Solution

Solution:- (C) Adiabatic process: $$\Delta U=-w$$

For:
Cyclic process $$:\Delta U=0 \Rightarrow q=-w$$

Isothermal process $$:\Delta U=0 \Rightarrow q=-w$$

Adiabatic process: $$q=0 \Rightarrow \Delta U=w$$

Isochoric process : $$w=0 \Rightarrow \Delta U=q$$
Question 2
1 / -0

Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below $$\displaystyle \frac{1}{2}\mathrm{C}1_{2}(\mathrm{g})$$ to $$\mathrm{C}1^{-}(\mathrm{a}\mathrm{q})$$
(using the data, $$\Delta_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}}\mathrm{H}_{\mathrm{C}1_{2}}^{\Theta}=240\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1},\ \Delta_{\mathrm{e}\mathrm{g}}\mathrm{H}_{\mathrm{C}1}^{\Theta}=-349\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1},\ \Delta_{\mathrm{h}\mathrm{y}\mathrm{d}}\mathrm{H}_{\mathrm{C}1^{-}}=-381\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$) will be

A
$$-850\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$

B
$$+120\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$

C
$$+152\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$

D
$$-610\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$

Solution

Oxidising power of chlorine in aqueous solution can be estimated by calculating the overall enthalpy change of reaction.
The reaction takes place with following steps:
$$1/2Cl_{2} -> Cl$$   ΔH1
$$Cl -> Cl^{-}$$    ΔH2
$$Cl^{-} -> Cl^{-} (aq) $$   ΔH3
ΔH will be the sum of the energy involved in these
steps.
Hence, ΔH = ΔH1 +ΔH2 +ΔH3
ΔH1 = 240/2 = 120kJ/mol, ΔH2 = -349kJ/mol, ΔH3 = -381kJ/mol
Hence, ΔH = 120-349-381 = -610kJ/mol
Question 3
1 / -0

Which of the following parameters does not characterize the thermodynamic state of matter?

A
temperature

B
pressure

C
work

D
volume

Solution
Question 4
1 / -0

The combustions of benzene (l) gives $$CO_2(g)$$ and $$H_2O(l)$$. Given that heat of combustion of benzene at constant volume is $$-3263.9 \ kJ\, mol^{-1}$$ at $$25^oC$$, heat of combustion (in $$kJ\,mol^{-1}$$) of benzene at constant pressure will be:
$$(R =8.314JK^{-1}\,mol^{-1})$$

A
$$+3260$$

B
$$-3267.6$$

C
$$+4152.6$$

D
$$-452.46$$

Solution

$$C_6H_6(l) + \dfrac{15}{2} O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$$

$$\Delta H = \Delta U + \Delta n RT$$

$$=-3263.9 + \dfrac {( 6- 7.5) \times 8.314 \times 298}{1000}$$

$$=-3263.9 - 3.7 $$

$$= -3267.6 \ kJ$$

Hence, option $$B$$ is correct.
Question 5
1 / -0

The standard Gibbs energy for the given cell reaction in $$kJ$$ $${mol}^{-1}$$ at $$298K$$ is:$$Zn(s)+{Cu}^{2+}(aq)\rightarrow {Zn}^{2+}(aq)+Cu(s)$$

$${E}^{o}_{cell}=2V$$ at $$298K$$
(Faraday's constant, $$F=96000\ C{mol}^{-1}$$)

A
$$-384$$

B
$$-192$$

C
$$192$$

D
$$384$$

Solution

We know,
$$\Delta G^o = -nFE^o_{cell} $$

Since 2 electrons are involved so $$n=2$$.

$$\Delta G = -nFE^o_{cell} = -2 \times 96000 \times 2 =-384000\ J= -384\ kJ$$
Question 6
1 / -0

The INCORRECT match in the following is :

A
$$\Delta G^0 < 0,K< 1$$

B
$$\Delta G^0 < 0,K= 1$$

C
$$\Delta G^0 > 0,K< 1$$

D
$$\Delta G^0 < 0,K> 1$$

Solution

We know,
$$\Delta G=\Delta G^0+RT\ lnK$$

At equilibrium,

$$\Delta G=\Delta G^0+RT\ lnK=0$$

$$\Rightarrow \Delta G^0=−RT\ lnK$$

$$−RT\ lnK<0$$

$$RT\ lnK>0$$

$$lnK>0$$

$$\Rightarrow K>1$$
Question 7
1 / -0

An ideal gas undergoes a cyclic process as shown in Figure.

$$\Delta U_{BC}=-5$$ kJ $$mol^{-1}$$, $$q_{AB}=2$$ kJ $$mol^{-1}$$

$$W_{AB}=-5$$ kJ $$mol^{-1}$$, $$W_{CA}=3$$ kJ $$mol^{-1}$$

Heat absorbed by the system during process CA is:

A
$$+18$$ kJ $$mol^{-1}$$

B
$$-18$$ kJ $$mol^{-1}$$

C
$$-5$$ kJ $$mol^{-1}$$

D
$$+5$$ kJ $$mol^{-1}$$

Solution

$$\Delta U _{AB} = q_{AB}+W_{AB}=2+(-5)=-3 kJ/mol$$

$$\Delta U _{BC} = -5 kJ/mol$$

For cyclic process, $$\Delta U = 0$$

$$\Delta U _{AB}+ \Delta U _{BC} + \Delta U_{CA} = 0$$

$$ \Delta U_{CA} = -\Delta U _{AB}- \Delta U _{BC} $$

$$ \Delta U_{CA} = - (-3 )- (-5)=8 kJ/mol$$

$$ \Delta U_{CA} =q_{CA}+W_{CA} $$

$$ 8 =q_{CA}+3$$

$$ q_{CA}=+5kJ/mol$$

Heat absorbed has positive sign.
Question 8
1 / -0

A process will be spontaneous to all temperatures if:

A
$$\Delta H > 0$$ and $$\Delta S < 0$$

B
$$\Delta H < 0$$ and $$\Delta S > 0$$

C
$$\Delta H > 0$$ and $$\Delta S > 0$$

D
$$\Delta H < 0$$ and $$\Delta S < 0$$

Solution

Solution:- (B) $$\Delta H < 0$$ and $$\Delta S > 0$$

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

For reaction to be spontaneous process at all temperature, $$\Delta G < 0$$ and it is possible when $$\Delta H < 0$$ and $$\Delta S > 0$$ because it gives $$\Delta G < 0$$.
Question 9
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Compute the heat of formation of liquid methyl alcohol (in kJ mol $$^{ -1 }$$) using the following data. The heat of vaporisation of liquid methyl alcohol is 38 kJ mol$$^{ -1 }$$.

The heat of formation of gaseous atoms from the elements in their standard states:
$$H =$$ 218 kJ mol$$^{ -1 }$$., $$C = 715$$ kJ mol$$^{ -1 }$$. , $$O = 249$$ kJ mol$$^{ -1 }$$.

Average bond energies:
$$C-H = 415$$ kJ mol$$^{ -1 }$$.
$$C-O = 356$$ kJ mol$$^{ -1 }$$.
$$O-H = 463$$ kJ mol$$^{ -1 }$$.

A
$$\Delta H = -266$$ kJ mol$$^{ -1 }$$.

B
$$\Delta H = +266$$ kJ mol$$^{ -1 }$$.

C
$$\Delta H = -190$$ kJ mol$$^{ -1 }$$.

D
None of these

Solution

$$CH_3OH_{(l)}\longrightarrow CH_3OH_{(g)}; \Delta H = 38 $$ kJ mol$$^{ -1 }$$.   .....(1)

$$\dfrac{ 1 }{ 2 }$$ $${H_2}_{(g)}$$ $$\longrightarrow H_{(g)}; \Delta H = 218 $$ kJ mol$$^{ -1 }$$.   .....(2)

$$C(graphite)\longrightarrow C_{(g)}; \Delta H = 715 $$ kJ mol$$^{ -1 }$$.   .....(3)

$$\frac{ 1 }{ 2 }{O_2}_{(g)}\longrightarrow O_{(g)}; \Delta H = 249 $$ kJ mol$$^{ -1 }$$.   .....(4)

$$C-H_{(g)}\longrightarrow C_{(g)} + H_{(g)}; \Delta H = 415 $$ kJ mol$$^{ -1 }$$.   .....(5)

$$C-O_{(g)}\longrightarrow C_{(g)} + O_{(g)}; \Delta H= 356 $$ kJ mol$$^{ -1 }$$.   .....(6)

$$O-H_{(g)}\longrightarrow O_{(g)} + H_{(g)}; \Delta H = 463 $$ kJ mol$$^{ -1 }$$.   .....(7)

$$C(s) + 2{H_2}_{(g)} + \frac{ 1 }{ 2 }{O_2}_{(g)}\longrightarrow CH_3OH_{(g)}$$ .....(8)

The following energies are required:

$$2{H_2}_{(g)}\longrightarrow 4H_{(g)}$$

$$C(graphite)\longrightarrow C_{(g)}$$

$$\frac{1 }{ 2 }{O_2}_{(g)}\longrightarrow O_{(g)}$$

The total energy required is

$$(218\times 4) + 175 + 249 = 1836\ kJ$$ .....From equation (5), (6) and (7)

The following bonds are formed:

Three C-H bonds, one C-O bond, and O-H bond

The total energy released is

$$(3\times 415) + 356 + 463 = 2064\ kJ$$

Net energy released for $$CH_3OH_{(l)} = 228 + 38 = 266\ kJ$$ .... From eq (1) and (2)

Therefore,

$$\Delta H_fCH_3OH_{(l)} = -266$$ kJ mol$$^{ -1 }$$

Hence, the correct option is $$A$$
Question 10
1 / -0

The $$\Delta_f H^{ \ominus }$$ for $$CO_2(g), CO(g) and H_2O(g)$$ are $$-393.5, -110.5 and -214.8 kJ mol^{ -1 }$$, respectively. The standard enthalpy change $$(in kJ mol^{ -1 })$$ for the reaction $$CO_2(g) + H_2(g)\longrightarrow CO(g) + H_2O(g)$$ is

A
$$524.1$$

B
$$+41.2$$

C
$$-262.5$$

D
$$-41.2$$

Solution

$$CO_2(g) + H_2(g)\longrightarrow CO(g) + H_2O(g)$$
$$\Delta H^{

\ominus } = \left\{ \left[ \Delta _{ f }H^{ \ominus }CO,(g)+\Delta _{ f

}H^{ \ominus }H_{ 2 }O,(g) \right] -\Delta _{ f }H^{ \ominus }CO_{ 2

},(g) \right\}$$
$$\quad\quad = \left[ (-110.5)+(-241.8)-(-393.5+0) \right] $$
$$\quad\quad = +41.2 kJ$$
$$\Delta_fH^{ \ominus }H_2, (g) = 0$$
$$\Delta_fH^{ \ominus } = 0$$ in elementary state.

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Thermodynamics Test - 9

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Thermodynamics Test - 9

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Question 1

Cylic process $$:q=-w$$

Isothermal process $$:q=-w$$

Adiabatic process: $$\Delta U=-w$$

Isochoric process: $$\Delta U=q$$

Solution

Question 2

$$-850\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$

$$+120\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$

$$+152\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$

$$-610\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$

Solution

Question 3

temperature

pressure

work

volume

Solution

Question 4

$$+3260$$

$$-3267.6$$

$$+4152.6$$

$$-452.46$$

Solution

Question 5

$$-384$$

$$-192$$

$$192$$

$$384$$

Solution

Question 6

$$\Delta G^0 < 0,K< 1$$

$$\Delta G^0 < 0,K= 1$$

$$\Delta G^0 > 0,K< 1$$

$$\Delta G^0 < 0,K> 1$$

Solution

Question 7

$$+18$$ kJ $$mol^{-1}$$

$$-18$$ kJ $$mol^{-1}$$

$$-5$$ kJ $$mol^{-1}$$

$$+5$$ kJ $$mol^{-1}$$

Solution

Question 8

$$\Delta H > 0$$ and $$\Delta S < 0$$

$$\Delta H < 0$$ and $$\Delta S > 0$$

$$\Delta H > 0$$ and $$\Delta S > 0$$

$$\Delta H < 0$$ and $$\Delta S < 0$$

Solution

Question 9

$$\Delta H = -266$$ kJ mol$$^{ -1 }$$.

$$\Delta H = +266$$ kJ mol$$^{ -1 }$$.

$$\Delta H = -190$$ kJ mol$$^{ -1 }$$.

None of these

Solution

Question 10

$$524.1$$

$$+41.2$$

$$-262.5$$

$$-41.2$$

Solution

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