Thermodynamics Test

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Thermodynamics Test - 52

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Question 1
1 / -0

$$\Delta_f H^{ \ominus }$$ of hypothetical $$MgCl$$ is $$-125 kJ mol^{ -1 }$$ and for $$MgCl_2$$ is $$-642 kJ mol^{ -1 }$$. The enthalpy of disproportionation of $$MgCl$$ is $$-49x$$. Find the value of x.

A
$$x = 8$$

B
$$x = 8.5$$

C
$$x = 16$$

D
None of these

Solution

The reaction for formation of MgCl is as shown.
$$Mg(s) + 1/2Cl_2(g)\longrightarrow MgCl$$......(i) $$\Delta H_1 = -125 kJ mol^{ -1 }$$
The reaction for formation of $$MgCl_2$$ is as shown
$$Mg(s) + Cl_2(g)\longrightarrow MgCl_2$$......(ii) $$\Delta H_2 = -642 kJ mol^{ -1 }$$
The disproportionation reaction is as shown.
$$2MgCl\longrightarrow Mg + MgCl_2$$ ......(iii) $$\Delta H = ?$$
The reaction (i) is multiplied by 2 and subtracted from reaction (ii) to obtain reaction (iii)
The enthalpy of disproportionation of MgCl is
$$\Delta H = \Delta H_2 - 2\Delta H_1 = -642 - (2\times -125) = -392 kJ mol^{ -1 }$$
$$\therefore -49x = -392$$
$$x = 8$$
Question 2
1 / -0

Calculate $$\Delta_f H^{ \ominus }ICl(g)$$ from the data
$$\Delta H_{dissociation}Cl_2(g) = 57.9 Kcal mol^{ -1 }$$
$$\Delta H_{dissociation}I_2(g) = 36.1 Kcal mol^{ -1 }$$
$$\Delta H_{dissociation}ICl(g) = 50.5 Kcal mol^{ -1 }$$
$$\Delta H_{dissociation}I_2(g) = 15.0 Kcal mol^{ -1 }$$

A
$$\Delta H = 4 Kcal mol^{ -1 }$$

B
$$\Delta H = -4 Kcal mol^{ -1 }$$

C
$$\Delta H = 16 Kcal mol^{ -1 }$$

D
None of these

Solution

The balanced chemical equation for the formation of $$ICl$$ is as shown.
$$\frac { 1 }{ 2 }Cl_2(g) + \frac { 1 }{ 2 }I_2(s)\longrightarrow ICl(g) ..... \Delta_f H = ?$$
The balanced chemical equations for the dissociation and sublimation are as shown.
$$\begin{matrix} 1/2 \\ 1/2 \\ 1/2 \end{matrix}\begin{cases} I_{ 2 }(g)\longrightarrow 2I(g);\quad \Delta H_{ 1 }=36.1\times 1/2 \\ Cl_{ 2 }(g)\longrightarrow 2Cl(g);\Delta H_{ 2 }=57.9\times 1/2 \\ I_{ 2 }(s)\longrightarrow I_{ 2 }(g)+Cl(g);\Delta H_{ 3 }=15.0\times 1/2 \\ ICl(g)\longrightarrow I(g)+Cl(g);\Delta H_{ 4 }=50.5 \end{cases}$$

The enthalpy change for the formation of $$ICl$$(g) is
$$\Delta H = (\Delta H_1 + \Delta H_2 + \Delta H_3) - \Delta H_4$$
$$= \frac { 1 }{ 2 }(36.1 + 57.9 + 15.0) - 50.5$$
$$ = 4 Kcal mol^{ -1 }$$
Question 3
1 / -0

Using the enthalpies of formation, calculate the energy (kJ) released when $$3.00 g$$ of $$N{ H }_{ 3\left( g \right) }$$ reacts according to the following equation?
(Atomic weights: $$N = 14.00, H = 1.008$$).
$$4N{ H }_{ 3 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 4NO\left( g \right) +6{ H }_{ 2 }O\left( g \right) $$
$$\Delta HN{ H }_{ 3 }\left( g \right) =-46.1   { kJ }/{ mole }$$
$$\Delta HNO\left( g \right) =+90.2   { kJ }/{ mole }$$
$$\Delta H{ H }_{ 2 }O\left( g \right) =-241.8   { kJ }/{ mole }$$

A
$$34.3$$

B
$$30.8$$

C
$$39.9$$

D
$$42.6$$

Solution

Energy released, when four mole of $$NH_{ 3 }$$ reacts, is given by $$\displaystyle \Delta_{r}H = H_{ Products } - H_{ Reactants }$$ $$\displaystyle \Delta_{r}H = (4 \times \Delta HNO\left( g \right) + 6 \times \Delta H{ H }_{ 2 }O\left( g \right)) - (4 \times \Delta HN{ H }_{ 3 }\left( g \right) + 5 \times \Delta H{ O }_{ 2 }\left( g \right))$$
$$\displaystyle \Delta_{r}H = 4 \times 90.2 + 6 \times (-241.8) - 4 \times (-46.1) + 5 \times 0$$ (As $$\Delta H{ O }_{ 2 }\left( g \right) = 0$$) $$\displaystyle \Delta_{r}H = -905.6\ kJ$$ Thus, for four mole of $$NH_{ 3 }$$, i.e., $$(4\times14=) 68g$$ of $$NH_{ 3 }$$ amount of energy released is 905.6 kJ. Therefore, for 3g of $$NH_{ 3 }$$ amount of energy released will be $$\dfrac{(905.16 \times 3)}{68} = 39.9\ kJ$$
Question 4
1 / -0

a) Calculate heat of dissociation for Acetic acid from the following data:

$$CH_3COOH + NaOH\longrightarrow CH_3COONa + H_2O .... \Delta H = -13.2 Kcal$$

$$H^{ \oplus } + \overset { \ominus }{ O }H\longrightarrow H_2O; .... \Delta H = -13.7 Kcal$$.

b) calculate heat of dissociation for $$NH_4OH$$ if

$$HCl + NH_4OH\longrightarrow NH_4Cl + H_2O; \Delta H = -12.27 Kcal$$.

A
a) $$ 0.5\ kcal$$ , b)$$ 1.43\ cal$$

B
a) $$ -0.5\ kcal$$ , b)$$ -1.43\ cal$$

C
a) $$ 0.0\ kcal$$ , b)$$ 2.86\ cal$$

D
None of these

Solution

$$\Delta H\quad =\quad \Delta { H }_{ products }-{ \Delta H }_{ reactants }$$

There is no enthalpy during the formation of salt.

$$CH_3COOH + NaOH\longrightarrow CH_3COONa + H_2O .... \Delta H = -13.2\ kcal$$

$$H^{ \oplus } + \overset { \ominus }{ O }H\longrightarrow H_2O; .... \Delta H = -13.7\ Kcal$$

$$-13.7 = -13.2 - (\Delta { H }_{ C{ H }_{ 3 }COOH })$$

$$\Delta { H }_{ C{ H }_{ 3 }COOH } = 13.7- (-13.2) = -0.5$$ kCal

Thus, the heat of dissociation for Acetic acid= -0.5 kcal

$$HCl + NH_4OH\longrightarrow NH_4Cl + H_2O; \Delta H = -12.27\ kcal$$

$$-12.2 = -13.2 - (\Delta { H }_{ N{ H }_{ 4 }OH })$$

$$\Delta { H }_{ N{ H }_{ 4 }OH } = -13.7- (-12.27) = -1.43$$ kCal

Thus, the heat of dissociation for NH$$_4$$OH$$ = -$$1.43 cal

Above are enthalpies of formation. Negative of the above values give heat of dissociation.

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

$$\Delta_f H^{ \ominus }$$ of Cyclohexene $$(l)$$ and benzene at $$25^{ \circ}C$$ is $$-156$$ and $$+46 kJ mol^{ -1 }$$, respectively.
$$\Delta_{ hydrogenation }H^{ \ominus }$$ of Cyclohexene $$(l)$$ at $$25^{ \circ }C$$ is $$-119 kJ mol^{ -1 }$$.

Resonance energy of benzene is found to be $$-38x kJ mol^{ -1 }$$. Find the value of x.

A
4

B
2

C
8

D
None of these

Solution

This shows that the generation of one $$(C=C)$$ bond in Cychlohexene requires $$119 kJ mol^{ -1 }$$ of enthalpy. TO calculate RE resonance energy.

$$\Delta H^{ \ominus }_1 = \Delta H^{ \ominus }_2 = \Delta H^{ \ominus }_3 = 119 kJ mol^{ -1 }, \Delta H^{ \ominus }_4 = RE$$
$$\Delta H^{ \ominus }_5 = \Delta_f H^{ \ominus }(benzene) - \Delta_f H^{ \ominus }(cyclohexene)$$
$$ = 46 - (-156) = 205 kJ mol^{ -1 }$$
$$\therefore\quad From\quad Hess`\quad law:$$
$$\Delta H^{ \ominus }_4 = \Delta H^{ \ominus }_5 - (\Delta H^{ \ominus }_1 + \Delta H^{ \ominus }_2 + \Delta H^{ \ominus } _3)$$.
$$ = 205 - 3\times 119 = -152 kJ mol^{ -1 }$$
$$\therefore -38x = -152$$
$$x = 4$$
Question 6
1 / -0

For the reaction $$X_2O_4(l)\rightarrow 2XO_2(g)$$

$$\Delta U = 2.1\,Kcal, \Delta S = 20\,cal\,K^{ -1 }\,at\, 300\,K$$

Hence $$\Delta G$$ is:

A
$$+2.7\,Kcal$$

B
$$-2.7\,Kcal$$

C
$$+9.3\,Kcal$$

D
$$-9.3\,Kcal$$

Solution
Question 7
1 / -0

Determine $${ \Delta H }/{ kJ }$$ for the following reaction using the listed enthalpies of reaction:
$$4CO\left( g \right) +8{ H }_{ 2 }\left( g \right) \longrightarrow 3C{ H }_{ 4 }\left( g \right) +C{ O }_{ 2 }\left( g \right) +2{ H }_{ 2 }O\left( l \right) $$

$$C\left( graphite \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow CO\left( g \right) ; { \Delta H }/{ kJ }=-110.5kJ$$

$$CO\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right) ; { \Delta H }/{ kJ }=-282.9kJ$$

$${ H }_{ 2 }\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow { H }_{ 2 }O\left( l \right) ; { \Delta H }/{ kJ }=-285.8kJ$$

$$C\left( graphite \right) +2{ H }_{ 2 }\left( g \right) \longrightarrow C{ H }_{ 4 }\left( g \right) ; { \Delta H }/{ kJ }=-74.8kJ$$

A
-622.4

B
-686.2

C
-747.5

D
-653.5

Solution

$$4CO\left( g \right) +8{ H }_{ 2 }\left( g \right) \longrightarrow 3C{ H }_{ 4 }\left( g \right) +C{ O }_{ 2 }\left( g \right) +2{ H }_{ 2 }O\left( l \right) $$

$$C\left( graphite \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow CO\left( g \right) ; { \Delta H }/{ kJ }=-110.5kJ$$ ---1

$$CO\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right) ; { \Delta H }/{ kJ }=-282.9kJ$$ ----2

$${ H }_{ 2 }\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow { H }_{ 2 }O\left( l \right) ; { \Delta H }/{ kJ }=-285.8kJ$$ ----- 3

$$C\left( graphite \right) +2{ H }_{ 2 }\left( g \right) \longrightarrow C{ H }_{ 4 }\left( g \right) ; { \Delta H }/{ kJ }=-74.8kJ$$ ----- 4

Applying Hess's law,
$$\Delta { H }_{ r } =3\times (-\Delta H_1)+\Delta H_2+2\times \Delta H_3+3\times\Delta H_4$$

$$\Delta { H }_{ r } = \left[ 3\left( 110.5 \right) - 282.9 + 2\left( -285.8 \right) + 3\left( -74.8 \right) \right] $$

$$\Delta { H }_{ r } = -747.4\ kJ$$
Question 8
1 / -0

Use the given standard enthalpies of formation to determine the heat of reaction of the following reaction:
$$2LiOH+C{ O }_{ 2 }\left( g \right) \longrightarrow { Li }_{ 2 }C{ O }_{ 3 }\left( s \right) +{ H }_{ 2 }O\ (l)$$
$$\Delta { H }_{ f }^{ }LiOH\left( s \right) =-487.23{ kJ }/{ mole }$$
$$\Delta { H }_{ f }^{ }{ Li }_{ 2 }C{ O }_{ 3 }\left( s \right) =-1215.6{ kJ }/{ mole }$$
$$\Delta { H }_{ f }^{ }{ H }_{ 2 }O\ (l) =-285.85{ kJ }/{ mole }$$
$$\Delta { H }_{ f }^{ }C{ O }_{ 2 }\left( g \right) =-393.5{ kJ }/{ mole }$$

A
$$+303.4$$

B
$$-133.5$$

C
$$-198.6$$

D
$$+198.6$$

Solution

$$2LiOH+C{ O }_{ 2 }\left( g \right) \longrightarrow { Li }_{ 2 }C{ O }_{ 3 }\left( s \right) +{ H }_{ 2 }O\left( l\right) $$
$$\Delta { H }_{ f }^{ }LiOH\left( s \right) =-487.23\;{ kJ }/{ mole }$$
$$\Delta { H }_{ f }^{ }{ Li }_{ 2 }C{ O }_{ 3 }\left( s \right) =-1215.6\;{ kJ }/{ mole }$$
$$\Delta { H }_{ f }^{ }{ H }_{ 2 }O\left( l \right) =-285.85\;{ kJ }/{ mole }$$
$$\Delta { H }_{ f }^{ }C{ O }_{ 2 }\left( g \right) =-393.5\;{ kJ }/{ mole }$$
As we know,
Heat of reaction = Heat of formation of products - Heat of formation of reactants
$$={(-1215.6-285.85} )-\ {((2\times-487.23) +(-393.5))} = -133.5$$ kJ/mole
Question 9
1 / -0

Determine $$\Delta H$$ of the following reaction using the listed heats of formation:
$$4HN{ O }_{ 3 }\left( I \right) +{ P }_{ 4 }{ O }_{ 10 }\left( s \right) \longrightarrow 2{ N }_{ 2 }{ O }_{ 5 }\left( s \right) +4HP{ O }_{ 3 }\left( s \right) $$
$$\Delta { H }_{ f }^{ }HN{ O }_{ 3 }\left( I \right) =-174.1{ kJ }/{ mole }$$
$$\Delta { H }_{ f }^{ }{ N }_{ 2 }{ O }_{ 5 }\left( s \right) =-43.1{ kJ }/{ mole }$$
$$\Delta { H }_{ f }^{ }{ P }_{ 4 }{ O }_{ 10 }\left( s \right) =-2984.0{ kJ }/{ mole }$$
$$\Delta { H }_{ f }^{ }HP{ O }_{ 3 }\left( s \right) =-948.5{ kJ }/mole$$

A
-176.3

B
-199.8

C
+276.2

D
-242.4

Solution

The enthalpy change for the reaction is the difference between the enthalpy of products and the enthalpy of reactants.
$$\Delta { H }_{ r }=\left[ 2{ \left( \Delta { H }_{ f } \right) }_{ { N }_{ 2 }{ O }_{ 5 } }+4{ \left( \Delta { H }_{ f } \right) }_{ HP{ O }_{ 3 } }-4{ \left( \Delta { H }_{ 5 } \right) }_{ HN{ O }_{ 3 } }-{ \left( \Delta { H }_{ f } \right) }_{ { P }_{ 4 }{ O }_{ 10 } } \right] $$
$$\Delta { H }_{ r }=\left[ 2\left( -43.1 \right) +4\left( -948.5 \right) -4\left( -174.1 \right) -\left( -2984.0 \right) \right] $$
$$=-199.8 \: kJ/mol$$
Hence, the enthalpy change for the reaction is -199.8 kJ/mol.
Question 10
1 / -0

Calculate the value of $${ \Delta H }{\ ( kJ) }$$ for the following reaction using the listed thermochemical equations $$2C(s) + H_2(g) \rightarrow C_2H_ 2 (g)$$.

$$2{ C }_{ 2 }{ H }_{ 2 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 4C{ O }_{ 2 }\left( g \right) +2{ H }_{ 2 }O\left( l \right)$$ ; $$\Delta H^o= -2600\ kJ$$
$$C\left( s \right) +{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right)$$ ; $$\Delta H^o=-390\ kJ$$
$$2{ H }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ H }_{ 2 }O\left( l \right)$$ ; $$\Delta H^o =-572\ kJ$$

A
$$+184$$

B
$$+214$$

C
$$+202$$

D
$$+234$$

Solution

$$2{ C }_{ 2 }{ H }_{ 2 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 4C{ O }_{ 2 }\left( g \right) +2{ H }_{ 2 }O\left( l \right)$$ ; $$\Delta H^o_1= -2600\ kJ$$

$$C\left( s \right) +{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right)$$ ; $$\Delta H^o_2=-390\ kJ$$

$$2{ H }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ H }_{ 2 }O\left( l \right)$$ ; $$\Delta H^o_3 =-572\ kJ$$
Now,

$$\Delta { H }_{ r }=\left[ -\displaystyle\frac { 1 }{ 2 } { \left( \Delta { H }_{ 1 } \right) }+2{ \left( \Delta { H }_{ 2 } \right) }+\displaystyle\frac { 1 }{ 2 } { \left( \Delta { H }_{ 3} \right) } \right] $$

$$\Delta { H }_{ r }=-\displaystyle\frac { 1 }{ 2 } \left( -2600 \right) +2\left( -390 \right) -\displaystyle\frac { 1 }{ 2 } \times (-572)$$

$$\Delta { H }_{ r }=234\ kJ$$

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Thermodynamics Test - 52

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Thermodynamics Test - 52

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Question 1

$$x = 8$$

$$x = 8.5$$

$$x = 16$$

None of these

Solution

Question 2

$$\Delta H = 4 Kcal mol^{ -1 }$$

$$\Delta H = -4 Kcal mol^{ -1 }$$

$$\Delta H = 16 Kcal mol^{ -1 }$$

None of these

Solution

Question 3

$$34.3$$

$$30.8$$

$$39.9$$

$$42.6$$

Solution

Question 4

a) $$ 0.5\ kcal$$ , b)$$ 1.43\ cal$$

a) $$ -0.5\ kcal$$ , b)$$ -1.43\ cal$$

a) $$ 0.0\ kcal$$ , b)$$ 2.86\ cal$$

None of these

Solution

Question 5

4

2

8

None of these

Solution

Question 6

$$+2.7\,Kcal$$

$$-2.7\,Kcal$$

$$+9.3\,Kcal$$

$$-9.3\,Kcal$$

Solution

Question 7

-622.4

-686.2

-747.5

-653.5

Solution

Question 8

$$+303.4$$

$$-133.5$$

$$-198.6$$

$$+198.6$$

Solution

Question 9

-176.3

-199.8

+276.2

-242.4

Solution

Question 10

$$+184$$

$$+214$$

$$+202$$

$$+234$$

Solution

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