Thermodynamics Test

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Thermodynamics Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

Select correct statement.

A
Both lattice energy and hydration energies decrease with ionic size.

B
Lattice energy can be calculated using Born-Haber cycle.

C
If the anion is larger compared to the cation, the lattice energy will remain almost constant within a particular group.

D
All the above are correct statements.

Solution

All the given statements are correct.
(A) Both lattice energy and hydration energies decrease with ionic size With large size, the charge is more dispersed. hence, the extent of hydration is smaller. Also the packing is inefficient.
(B) Lattice energy can be calculated using Born-Haber cycle The BornHaber cycle is an approach to analyze reaction energies. The cycle is concerned with the formation of an ionic compound from the reaction of a metal (often a Group I or Group II element) with a halogen.
(C) If the anion is larger compared to the cation, the lattice energy will remain almost constant within a particular group
Question 2
1 / -0

An ideal monoatomic gas undergoes a process in which its internal energy U and density $$\rho$$ vary as $$U\rho\, =\, constant.$$ The ratio of change in internal energy and the work done by the gas is

A
$$\displaystyle \frac{3}{2}$$

B
$$\displaystyle \frac{2}{3}$$

C
$$\displaystyle \frac{1}{3}$$

D
$$\displaystyle \frac{3}{5}$$

Solution

As we know that internal energy(U) of a monoatomic gas is given as $$\dfrac{{m}{v}^{2}}{2}$$
and work done (W)=F.S
where $$F=\dfrac{{mn}{v}^{2}}{3Ln}$$
and $$S=L$$
so $$\dfrac{internal energy}{work done}$$=$$\dfrac{3}{2}$$
Question 3
1 / -0

Heat of reaction for the reaction is:

$$PCl_5(g)+H_2O(g)\longrightarrow POCl_3(g)+2HCl(g)$$

Given that,
$$P_{white} + (3/2)Cl_2(g) +\frac{1}{2}O_2(g)\longrightarrow POCl_3; \: \Delta H=-135.5 \:kcal$$

$$H_2( g ) + Cl_2( g )\longrightarrow +2HCl(g); \:                 \Delta H=-44.1 \:kcal$$

$$P(w)+(5/2)Cl_2(g)\longrightarrow PCl_5(g); \:    \Delta H=-89.6 \:kcal$$

$$H_2(g)+\frac{1}{2}O_2(g)\longrightarrow H_2O(g);\:                  \Delta H=-57.8\:kcal$$

A
$$-32.2 kcal$$

B
$$-52.5 kcal$$

C
$$-45.2 kcal$$

D
None of these

Solution

$$P_{white} + \dfrac 32 Cl_2(g) +\dfrac{1}{2}O_2(g)\longrightarrow POCl_3; \: \Delta H_1=-135.5 \:kcal$$

$$H_2( g ) + Cl_2( g )\longrightarrow +2HCl(g); \:                 \Delta H_2=-44.1 \:kcal$$

$$P(w)+ \dfrac52 Cl_2(g)\longrightarrow PCl_5(g); \:    \Delta H_3=-89.6 \:kcal$$

$$H_2(g)+\cfrac{1}{2}O_2(g)\longrightarrow H_2O(g);\:                  \Delta H_4=-57.8\:kcal$$

The final reaction is obtained by reversing 3 and 4 and adding 1 and 2 to it:

$$PCl_5(g)+H_2O(g)\longrightarrow POCl_3(g)+2HCl(g)$$

$$\Delta H = \Delta H_1 + \Delta H_2 - \Delta H_3 - \Delta H_4$$.

$$\implies$$ $$ \Delta H = - 135.5 - 44.1 + 89.6 + 57.8$$

$$ =- 32.2 kcal $$
Question 4
1 / -0

The dissolution of 1 mole of NaOH(s) in $$100 \: mole \: of \: H_2O(l)$$ give rise to evolution of heat as -42.34 kJ. However, if1 mole of NaOH(s) is dissolved in $$1000 \:mole \:of \:H_2O(l)$$ the heat given out is 42.76 kJ.

If the enthalpy change when $$900 \:mole \:of \:H_2O(l) $$ are added to a solution containing 1 mole of NaOH(s) in $$100 \:mole \:of \:H_2O$$ is $$x$$ $$kJ$$:

A
$$-0.42$$

B
$$0.42$$

C
$$0.24$$

D
None of these

Solution

$$\Delta H_1 for \: 1 \: mole \: NaOH \: in \: 100 \: mole \: H_2O(l) = -42.34 \: kJ$$

$$NaOH(s) + 100H_2O\longrightarrow NaOH(100H_2O); \: \Delta H= -42.34 \: kJ$$ ... (i)

$$NaOH(s) + 1000H_2O\longrightarrow NaOH(1000H_2O); \: \Delta H= -42.76 \: kJ$$ ... (ii)

By eqs. [(ii) - (i)]

$$NaOH(s) + 900H_2O\longrightarrow NaOH(900H_2O); \: \Delta H= -0.42 \: kJ$$

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

The heat measured for a reaction in bomb calorimeter is :

A
$$\Delta G$$

B
$$\Delta H$$

C
$$\Delta U$$

D
$$P\Delta V$$

Solution

A bomb calorimeter is a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction.

Hence, the heat measured for a reaction in bomb calorimeter is $$\Delta U$$.
Question 6
1 / -0

1 gram sample of $$NH_4NO_3$$ is decomposed in a bomb calorimeter. The temperature of the calorimeter increases by 6.12 K. The heat capacity of the system is 1.23 kJ/g-deg. What is the molar heat of decomposition for $$NH_4NO_3$$?

A
-7.53 kJ/mol

B
-398.1 kJ/mol

C
-16.1 kJ/mol

D
602 kJ/mol

Solution

The expression for the heat evolved is $$q= mS\Delta T$$

When 1g of ammonium nitrate decomposes, the heat evolved is $$q=1\times 1.23\times 6.12$$

When 1 mole of ammonium nitrate decomposes, the heat evolved is $$q=1\times 1.23\times 6.12\times 80=6 02.21 \:kJ/mol$$
Question 7
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The $$\Delta_f H^{\circ}$$ for $$CO_2 (g), \: CO(g) \: and \: H_2O(g)$$ are $$-393.5, -110.5$$ and $$-241.8 \: kJ \: mol^{-1}$$ respectively. The standard enthalpy change (in kJ) for the reaction is:

$$CO_2 (g)+ H_2 (g)\longrightarrow CO(g)+ H_2O(g )$$

A
524.1

B
41.2

C
-262.5

D
-41.2

Solution

First write the balanced chemical reactions for the formation of carbon dioxide, carbon monoxide and water.

$$C+ O_2 \longrightarrow CO_2; \:    \Delta H^{\circ}= -393.5 \:kJ$$ ... (i)

$$C+\frac{1}{2}O_2\longrightarrow CO;\:          \Delta H^{\circ}= -110.5 \:kJ$$     ...(ii)

$$H_2+\frac{1}{2}O_2\longrightarrow H_2O;\: \Delta H^{\circ}= -241.8 \:kJ$$     ...(iii)

Then obtain the required equation by equating $$(ii) + (iii) - (i):$$

$$CO_2+H_2\longrightarrow CO+H_2O; \:          \Delta H^{\circ}=+41.2$$

Hence, the standard enthalpy change for the reaction is 41.2 kJ/mol.
Question 8
1 / -0

The commercial production of water gas utilize the reaction under standard conditions:

$$C+H_2O(g)\longrightarrow H_2(g)+CO$$

The heat required for this endothermic reaction may be supplied by adding a limited amount of air and burning some carbon to $$CO_2$$. How many g of carbon must be burnt to $$CO_2$$ to provide enough heat for the water-gas conversion of 100 g carbon?

Neglect all heat losses to the environment. Also $$\Delta _fH^{\circ}$$ of CO, $$H_2O(g)$$ and $$CO_2$$ are -110.53, 241.81 and -393.51 kJ/mol respectively.

A
$$33.4 g$$

B
$$66.72 g$$

C
$$15.18 g$$

D
None of these

Solution

$$C+{ H }_{ 2 }O\left( g \right)\rightarrow { H }_{ 2 }+CO$$

$$\Delta { H }=\Delta { { H }_{ f } }\left( CO \right) -\Delta { { H }_{ f } }\left( { H }_{ 2 }O \right) $$

$$=\left\{ \left( -110.53 \right) -\left( -241.81 \right) \right\} kJ/mole=131.28kJ/mole$$

so, total amount of energy to be provided for conversion of 100 gm of C

$$\displaystyle(100gC)\left( \frac { 1mole C }{ 12.0gC } \right) \left( \frac { 131.28kJ }{ mol } \right) =1094kJ$$

To provide that energy, amount of C to be burnt

$$\displaystyle\left( 1094 \right) kJ\left( \frac { 1mole C }{ 393.51kJ } \right) \left( \frac { 12.0g C }{ mole C } \right) =33.4g$$
Question 9
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The enthalpies of formation of $$N_2O$$ and $$NO$$ are $$28$$ and $$90\ kJ mol^{-1}$$ respectively. The enthalpy of the reaction, $$2N_2O(g) +O_2(g)\longrightarrow 4NO(g)$$ is equal to :

A
8 kJ

B
88 kJ

C
-16 kJ

D
304 kJ

Solution

First write the balanced chemical; equations for he formation of dinitrogen oxide and nitric oxide.

$$N_ 2+\frac{1}{2}O_2\longrightarrow N_2O; \:     \Delta H=28 \:kJ$$...(i)

$$\frac{1}{2}N_ 2+\frac{1}{2}O_2\longrightarrow NO; \:         \Delta H=90 \:kJ$$...(ii)

Then multiply second equation with 4 and first equation with 2.

$$2N_ 2+O_2\longrightarrow 2N_2O; \:     \Delta H=2 \times 28 \:kJ$$...(iii)

$$2N_ 2+2O_2\longrightarrow 4NO; \:         \Delta H=4 \times 90 \:kJ$$...(iv)

Now substract third equation from fourth equation.

$$2N_2O +O_2\longrightarrow 4NO; \:                        \Delta H=304 \:kJ$$.

Thus,the enthalpy of the reaction $$2N_2O(g) +O_2(g)\longrightarrow 4NO(g)$$ is 304 kJ.
Question 10
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Which of the following has the same value as $$\Delta_f H^{ \ominus }$$ , $$CO$$
a. $$\frac { 1 }{ 2 } \Delta_f H^{ \ominus } (CO_2)$$
b. $$\frac { 1 }{ 2 } \Delta_c H^{ \ominus } (graphite)$$
c. $$\Delta_f H^{ \ominus } (CO_2) - \Delta_f H^{ \ominus } (graphite)$$
d. $$\Delta_c H^{ \ominus } (graphite) - \Delta _c H^{ \ominus } (CO)?$$

A
a

B
b

C
c

D
d

Solution

As we know,
$$C(graphite) + O_2(g) \rightarrow CO_2(g) ;\quad \Delta _cH =x\ kJ$$................1

$$CO + 1/2 O_2(g) \rightarrow CO_2(g) $$; $$\Delta_c H^{ \ominus }$$ $$(CO) = z\ kJ$$.......................2

Equation (1)-(2) gives (3) eqaution.

$$C(graphite) + 1/2 O_2(g) \rightarrow CO(g) $$; $$\Delta_f H^{ \ominus }{(CO)}=y\ kJ$$.......................3

From these equations, it can be seen that

$$\Delta_f H^{ \ominus }$$ $$(CO)$$ = $$\Delta_c H^{ \ominus } (graphite) - \Delta _c H^{ \ominus } (CO)$$

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Thermodynamics Test - 50

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Thermodynamics Test - 50

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SHARING IS CARING

Question 1

Both lattice energy and hydration energies decrease with ionic size.

Lattice energy can be calculated using Born-Haber cycle.

If the anion is larger compared to the cation, the lattice energy will remain almost constant within a particular group.

All the above are correct statements.

Solution

Question 2

$$\displaystyle \frac{3}{2}$$

$$\displaystyle \frac{2}{3}$$

$$\displaystyle \frac{1}{3}$$

$$\displaystyle \frac{3}{5}$$

Solution

Question 3

$$-32.2 kcal$$

$$-52.5 kcal$$

$$-45.2 kcal$$

None of these

Solution

Question 4

$$-0.42$$

$$0.42$$

$$0.24$$

None of these

Solution

Question 5

$$\Delta G$$

$$\Delta H$$

$$\Delta U$$

$$P\Delta V$$

Solution

Question 6

-7.53 kJ/mol

-398.1 kJ/mol

-16.1 kJ/mol

602 kJ/mol

Solution

Question 7

524.1

41.2

-262.5

-41.2

Solution

Question 8

$$33.4 g$$

$$66.72 g$$

$$15.18 g$$

None of these

Solution

Question 9

8 kJ

88 kJ

-16 kJ

304 kJ

Solution

Question 10

a

b

c

d

Solution

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