Thermodynamics Test

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Thermodynamics Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

Calculate $${ \Delta H }/{ kJ }$$ for the following reaction using the listed standard enthalpy of reaction data.
$$2{ N }_{ 2 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 2{ N }_{ 2 }{ O }_{ 5 }\left( s \right) $$

$${ N }_{ 2 }\left( g \right) +3{ O }_{ 2 }\left( g \right) +{ H }_{ 2 }\left( g \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) ; \ { \Delta H }/{ kJ }=-414.0$$

$${ N }_{ 2 }{ O }_{ 5 }\left( s \right) +{ H }_{ 2 }O\left( l \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) ; \ { \Delta H }/{ kJ }=-86.0$$

$$2{ H }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ H }_{ 2 }O\left( l \right) ; \ { \Delta H }/{ kJ }=-571.6$$

A
-84.4

B
-243.6

C
-71.2

D
-121.8

Solution

$$2{ N }_{ 2 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 2{ N }_{ 2 }{ O }_{ 5 }\left( s \right) $$ ----- a

Given that,
$${ N }_{ 2 }\left( g \right) +3{ O }_{ 2 }\left( g \right) +{ H }_{ 2 }\left( g \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) ; { \Delta H }/{ kJ }=-414.0$$ -------- 1

$${ N }_{ 2 }{ O }_{ 5 }\left( s \right) +{ H }_{ 2 }O\left( l \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) ; { \Delta H }/{ kJ }=-86.0$$ -------- 2

$$2{ H }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ H }_{ 2 }O\left( l \right) ; { \Delta H }/{ kJ }=-571.6$$ ------------- 3

Applying Hess's law,
Eqn (a) = 2 x Eqn (1) - 2 x Eqn (2) - Eqn (3)

$$\Delta { H }_{ r }=\left[ 2\left( -414 \right) +2\left( 86 \right) +571.6 \right] $$

$$\Delta { H }_{ r }=-84.4\ kJ$$
Question 2
1 / -0

Use the given standard enthalpies of formation to determine the heat of reaction of the following reaction:
$${ C }_{ 2 }{ H }_{ 5 }OH\left( l\right) +3{ O }_{ 2 }\left( g \right) \longrightarrow 2C{ O }_{ 2 }\left( g \right) +3{ H }_{ 2 }O\left( g \right) $$
$$\Delta { H }_{ f }^{ }{ C }_{ 2 }{ H }_{ 5 }OH\left( l \right) =-277.7{ kJ }/{ mole }$$
$$\Delta { H }_{ f }^{ }C{ O }_{ 2 }\left( g \right) =-393.5{ kJ }/{ mole }$$
$$\Delta { H }_{ f }^{ }{ H }_{ 2 }O\left( g \right) =-241.8{ kJ }/{ mole }$$

A
-1456.3

B
-1234.7

C
-1034.0

D
-1119.4

Solution

$$\Delta { H }_{ r }=\left[ 2{ \left( \Delta { H }_{ f } \right) }_{ C{ O }_{ 2 } }+3{ \left( \Delta { H }_{ f } \right) }_{ { H }_{ 2 }O }-{ \left( \Delta { H }_{ f } \right) }_{ { C }_{ 2 }{ H }_{ 5 }OH } \right] $$
$$\Delta { H }_{ r }=\left[ 2\left( -393.5 \right) -3\left( 241.8 \right) +277.7 \right] $$
$$\Delta { H }_{ r }=-1234.7{ kJ }/{ mole }$$
Question 3
1 / -0

0.16 g of methane was subjected to combustion at $$27^{\circ}C$$ in a bomb calorimeter. The temperature of the calorimeter system (including water) was found to rise by $$0.5^{\circ}C$$.
Calculate the heat of combustion of methane at constant pressure. The thermal capacity of the calorimeter system is $$17.7 \: kJ \: K^{-1}$$.
$$[R = 8.314 \: JK^{-1}\: mol^{-1}]$$.

A
-890

B
890

C
-750

D
None of these

Solution
Question 4
1 / -0

The enthalpy change for the following process at $${25}^{o}C$$ and under constant pressure at $$1$$ atm are as follows:
$${CH}_{4}(g)\longrightarrow C(g)+4H(g)$$ $${\Delta}_{r}H=396kcal/mole$$
$${C}_{2}{H}_{6}(g)\longrightarrow 2C(g)+6H(g)$$ $${\Delta}_{r}H=676kcal/mole$$

Calculate $$C-C$$ bond energy in $${C}_{2}{H}_{6}$$ and heat formation of $${C}_{2}{H}_{6}(g)$$.

[Given: $${\Delta}_{sub}C(s)=171.8kcal/mole$$ $$B.E(H-H)=104.1kcal/mole$$]

A
$$B.E(C-C)=82kcal/mol, {\Delta}_{f}H[{C}_{2}{H}_{6}(g)]=-20.1kcal/mol$$

B
$$B.E(C-C)=-82kcal/mol, {\Delta}_{f}H[{C}_{2}{H}_{6}(g)]=-20.1kcal/mol$$

C
$$B.E(C-C)=82kcal/mol, {\Delta}_{f}H[{C}_{2}{H}_{6}(g)]=+20.1kcal/mol$$

D
None of these

Solution

$$\displaystyle {\Delta}_{f}H[{C}_{2}{H}_{6}(g)]=-676+343.6+312.3=-676+655.9=-20.1kcal/mol$$

$$\displaystyle 4(B.E(C-H))=396 (B.E(C-C))+6(99)=676$$

$$\displaystyle B.E(C-C)=99\ (C-C)=676-594=82kcal/mol$$
Question 5
1 / -0

Use the given bond enthalpy data to estimate the $$\Delta H\left( kJ \right) $$ for the following reaction.
$$\left( C - H = 414 kJ, H - Cl = 431 kJ, Cl - Cl = 243 kJ, C - Cl = 331 kJ \right) $$.
$$C{ H }_{ 4 }\left( g \right) + 4{ Cl }_{ 2 }\left( g \right) \longrightarrow C{ Cl }_{ 4 }\left( g \right) + 4HCl\left( g \right) $$

A
620

B
330

C
420

D
105

Solution

The enthalpy change for the combustion reaction is the difference between the total bond enthalpy of reactants and the total bond enthalpy of products.

$$\Delta { H }_{ r }^{ }=\left[ 4\Delta { H }_{ C-H }+4\Delta { H }_{ Cl-Cl }-4\Delta { H }_{ C-Cl }-4\Delta { H }_{ H-Cl } \right] $$

Substittue values in the above expression.

$$\Delta { H }_{ r }^{ }=\left[ 4\times 414+4\times 243-4\times 331-4\times 431 \right] $$

$$\Delta { H }_{ r }^{ }=-420 \: kJ$$
Question 6
1 / -0

Caesium chloride is formed according to the following equation:

$$Cs(s)+0.5{Cl}_{2}(g)\longrightarrow CsCl(s)$$

The enthalpy of sublimation of $$Cs$$, enthalpy of dissociation of chlorine, ionization energy of $$Cs$$ and electron affinity of chlorine are $$81.2, 243.0, 375.7$$ and $$-348.3kJ$$ $${ol}^{-1}$$. The energy change involved in the formation of $$CsCl$$ is $$388.6\ kJ.{mol}^{-1}$$. Calculate the lattice energy of $$CsCl$$.

A
$$-618.7\ kJ{mol}^{-1}$$

B
$$+618.7\ kJ{mol}^{-1}$$

C
$$1315.2\ kJ{mol}^{-1}$$

D
None of these

Solution
Question 7
1 / -0

Using the data (all values are in $$kJ/mol$$ at $${25}^{o}C$$) given below:

(i) Enthalpy of polymerization of ethylene$$=-72$$
(ii) Enthalpy of formation of benzene $$(l)=49$$
(iii) Enthalpy of vaporization of benzene $$(l)=30$$
(iv) Resonance energy of benzene $$(l)=-152$$
(v) Heat of formation of gaseous atoms from the elements in their standard states $$H=218, C=715$$.

Average bond energy of $$C-H=415$$. Calculate the $$B.E$$ of $$C-C$$ and $$C=C$$.

A
$$C-C=343.67$$, $$C=C=615.33$$

B
$$C-C=615.33$$, $$C=C=342.67$$

C
$$C-C=343.67$$, $$C=C=959$$

D
None of these

Solution

Given
For the polymerization of ethylene,

$$n({CH}_{2}={CH}_{2})\rightarrow {(-{CH}_{2}-{CH}_{2}-)}_{n};$$ $$\Delta H=-72$$

i.e., $${B}_{C=C}-2{B}_{C-C}=-72\ \ .......(i)$$

For the formation of benzene

$$6{C}_{(s)}+3{H}_{2(g)}\longrightarrow {C}_{6}{H}_{6(l)};$$ $$\Delta H=49\ \ ......(ii)$$

For the vaporisation of benzene

$${C}_{6}{H}_{6(l)}\rightarrow {C}_{6}{H}_{6(g)}$$ $$\Delta H=30$$

R.E of $${C}_{6}{H}_{6}=-152$$

For the formation of H atom from hydrogen molecule in standard state,

$$\cfrac{1}{2} {H}_{2}\rightarrow H;$$ $$\Delta H=218$$

For the formation of gaseous C atom

$${C}_{(s)}\rightarrow {C}_{(g)};$$ $$\Delta H=715$$

The average bond energy $${B}_{C-H}=415$$ for equation (2)

$$(6\times 715+6\times 218)-(3{B}_{C-C}+3{B}_{C=C}+6\times 415-(-152))=49+30$$

$${B}_{C-C}+{B}_{C=C}=959..........(iii)$$

from equation $$(i)$$ and $$(iii)$$

The bond energy of $$C-C$$ bond $${B}_{C-C}=343.66$$ and the bond energy of C=C bond $${B}_{C}=C=615.33$$
Question 8
1 / -0

Use the given standard enthalpies of formation to determine the heat of reaction of the following reaction: $$TiCL_4(g)+2\:H_2O(g)\rightarrow TiO_2(g)+4\:HCl(g)$$

$$\Delta H^{\circ}_f\:TiCL_4(g)=-763.2\:kJ/mol$$
$$\Delta H^{\circ}_f\:TiO_2(g)=-944.7\:kJ/mol$$
$$\Delta H^{\circ}_f\:H_2O(g)=-241.8\:kJ/mol$$
$$\Delta H^{\circ}_f\:HCl(g)=-92.3\:kJ/mol$$

A
$$-\:278.1$$

B
$$+\:369.2$$

C
$$+\:67.1$$

D
$$-\:67.1$$

Solution

The enthalpy change for the reaction is the difference in the enthalpies of formation of products and the enthalpies of formation of reactants.

$$\Delta H_r=[(\Delta H_f)_{TiO_2}+4(\Delta H_f)_{HCl} - [(\Delta H_f)_{TiCl_4}+2(\Delta H_f)_{H_2O}]$$

Substitute values in the above expression.

$$\Delta H_r=-944.7 +(4\times -92.3) - [-763.2+(2\times -241.8)$$

$$\Delta H_r=-67.1\:kJ/mol$$

Hence, the heat of reaction for the hydrolysis of titanium tetrachloride is - 67.1 kJ/mol.
Question 9
1 / -0

A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0°C. As it does so, it absorbs 207 J of heat. The values of q and W for the process will be :
[Take : R = 8.314 J / mol K, In 7.5 = 2.01]

A
q = +207, W = -207

B
q = -207, W = -207

C
q = -207, W = +207

D
q = +207, W = +207

Solution
Question 10
1 / -0

A swimmer coming out from a pool is covered with a film of water weighing about 18 g. Calculate the internal energy of vaporisation at 100°C.
[$$\Delta H_{vap}^o$$ for water at 373 K = 40.66 kJ/mol ]

The correct option is:

A
35.67 kJ

B
36.64 kJ

C
37.56 kJ

D
None of the above

Solution