Thermodynamics Test

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Thermodynamics Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

Use the following data to calculate second electron ainity of oxygen, i.e., for the process
$$O^{-}(g) + e^{-}(g) \rightarrow O^{2-}(g)$$
Is the $$O^{2-}$$ ion stable in the gas phase?.Why is it stable in solid MgO?
Heat of sublimation of $$Mg(s) = + 147.7 kJ mol^{-1}$$
Ionisation energy of Mg(g) to form
$$Mg^{2+}(g) = + 2189.0 kJ mol^{-1}$$
Bond dissociation energy for $$O_2 = + 498.4 kJmol^{-1}$$
First electron affinity of $$O(g) = - 141.0 kJ mol^{-1}$$
Heat formation of $$MgO(s) = -601.7 kJ mol^{-1}$$
Lattice energy of $$MgO = -3791.0 kJ mol^{-1}$$

A
601.7

B
744.4

C
1346.1

D
147.7

Solution

Option (B) is correct.
$$\triangle H_1=-1346.1+q\\ Mg(s)+\frac { 1 }{ 2 } O_{ 2 }(g)\rightarrow MgO(s)\\ \triangle H_{ 2 }=-601.7\\ By\quad Born-Haber\quad Cycle(based\quad on\quad Hess\quad law)\\ \triangle H_{ 1 }=\triangle H_{ 2 }\\ -1346.1+q=-601.7\\ \qquad \qquad q=744.4kJ\quad mol^{-1}$$
Question 2
1 / -0

The energy change for the alternating reaction that yields chlorine sodium $$(Cl^{+}Na^{-})$$ will be:
$$2Na(s)\, +\, Cl_2(g)\,\rightarrow\, 2Cl^{+}Na^{-}(s)$$
Given that:
Lattice energy of $$NaCl\,=\,-787\, kJ\,mol^{-1}$$
Electron affinity of $$Na\,=\,-52.9\, kJ\, mol^{-1}$$
Ionisation energy of $$Cl\, =\, +\,1251\, kJ\, mol^{-1}$$
BE of $$Cl_2\,=\,244\, kJ\, mol^{-1}$$
Heat of sublimation of $$Na(s)\, =\,107.3\, kJ\, mol^{-1}$$
$$\Delta H_f(NaCl)\, =\,-411\, kJ\, mol^{-1}$$.

A
+640 kJ

B
+1280 kJ

C
-410 kJ

D
+410 kJ
Question 3
1 / -0

Determine $${ \Delta }{ U }^{ o }$$ at $$300K$$ for the following reaction using the listed enthalpies of reaction:

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

$$C_{(graphite)}+1/2{ O }_{ 2 }(g)\longrightarrow CO(g);\quad \Delta { { H }_{ 1 } }^{ o }=-110.5kJ$$

$$CO(g)+1/2{ O }_{ 2 }(g)\longrightarrow { CO }_{ 2 }(g);\quad \Delta { { H }_{ 2 } }^{ o }=-282.9kJ$$

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

$$C_{(graphite)}+2{ H }_{ 2 }(g)\longrightarrow { CH }_{ 4 }(g);\quad \Delta { { H }_{ 4 } }^{ o }=-74.8kJ$$

A
$$653.5\ kJ$$

B
$$-686.2\ kJ$$

C
$$-747.4\ kJ$$

D
None of these

Solution
Question 4
1 / -0

Using the listed information calculate $${ \Delta }_{ r }{ G }^{ o }$$ (in kJ/mol) at $${ 27 }^{ o }C$$.
$${ Co }_{ 3 }{ O }_{ 4 }(s)+4CO(g)\longrightarrow 3Co(s)+4{ CO }_{ 2 }(g)$$

Given : At $$300K$$ , $$\Delta { { H }^{ o } }_{ f }(kJ/mol) $$ are $$-891,\ -110.5,\ 0.0$$ and $$-393.5$$ respectively.
$${ S }^{ o }(J/K.mol)$$ are $$102.5,\ 197.7, \ 30.0$$ and $$213.7$$ respectively.

A
$$214.8$$

B
$$-195.0$$

C
$$-200.3$$

D
$$-256.45$$
Question 5
1 / -0

The lattice energy of NaCl(s) using the following data will be:
heat of sublimation of $$Na(s)\,=\,S$$
$$(IE)_1$$ of $$Na\,(g)\,=\,I$$
bond dissociation energy of $$Cl_2\,(g)\,=\,D$$
electron affinity of $$Cl\,(g)\,=\,-E$$
heat of formation of $$NaCl(s)\,=\,-Q$$

A
Lattice energy $$-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$$

B
Lattice energy $$-U\, =\, S\, -\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$$

C
Lattice energy $$-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, +\, E\, -\,Q$$

D
Lattice energy $$-U\, =\, S\, -\, I\, -\,\displaystyle \frac{D}{2}\, +\, E\, +\,Q$$
Solution
An estimate of the strength of the bonds in an ionic compound can be obtained by measuring the lattice energy of the compound, which is the energy given off when oppositely charged ions in the gas phase come together to form a solid.
The Lattice energy of NaCl from its elements Sodium and Chlorine in their stable forms is modeled in five steps in the diagram:
1. Enthalpy change of atomization enthalpy of lithium
2. Ionization enthalpy of lithium
3. Atomization enthalpy of fluorine
4. Electron affinity of fluorine
5. Lattice enthalpy
$$-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$$
Question 6
1 / -0

Fixed mass of an ideal gas contained in a $$24.63L$$ sealed rigid vessel at $$1$$ atm is heated from $$-{ 73 }^{ o }C$$ to $${ 27 }^{ o }C$$. Calculate change in Gibb's energy if entropy of gas is a function of temperature as $$S=2+{ 10 }^{ -2 }T(J/K)$$: (Use $$1$$ atm $$L=0.1kJ$$)

A
$$1231.5J$$

B
$$1281.5J$$

C
$$781.5J$$

D
$$0$$
Question 7
1 / -0

If the Gibbs free energy change when 1 mole of $$\displaystyle NaCl$$ is dissolved in water at 298K is $$x$$ kJ, then -$$1000x$$ is
Given that,
(a) Lattice energy of $$\displaystyle NaCl$$ = $$\displaystyle 778\ kJ{ mol }^{ -1 }$$
(b) Hydration energy of $$\displaystyle NaCl = -774.3\ kJ{ mol }^{ -1 }$$
(c) Entropy change at $$\displaystyle 298K=43\ J{ mol }^{ -1 }$$.

A
9117

B
9441

C
9114

D
9141

Solution

$$\displaystyle { \Delta H }_{ dissolution }={ \Delta H }_{ (ionisation) }+{ \Delta H }_{ (hydration) }$$$$\displaystyle $$$$\displaystyle =778-774.3$$

$$\displaystyle =3.7kJ{ mol }^{ -1 }=3700J{ mol }^{ -1 }$$

$$\displaystyle { \Delta S }_{ dissolution }=43J{ K }^{ -1 }$$

$$\displaystyle \therefore \Delta G_{ dissolution }=\Delta H-T\Delta S$$

$$\displaystyle =3700-298\times 43=-9114J$$

$$\displaystyle \Delta G=-9.114kJ$$
Question 8
1 / -0

For the reaction $$2Al + Fe_2O_3\longrightarrow 2Fe+Al_2O_3$$, the standard heat enthalpy of $$Fe_2O_3$$ and $$Al_2O_3$$ are $$-196.5$$ and $$ -399.1$$ kcal respectively. $$\Delta H^{\circ}$$ for the reaction is:

A
$$ -252.4 \: kcal$$

B
$$-135.5 \: kcal$$

C
$$ -202.6 \: kcal$$

D
none of the above

Solution

$$\Delta H^{\circ}_{Reaction}-\Delta H^{\circ}_{Products}-\Delta H^{\circ}_{Reactants}$$
$$=2\times \Delta H^{\circ}_{Fe}+\Delta H^{\circ}_{Al_2O_3}-[2\times \Delta H^{\circ}_{Al}+ \Delta H^{\circ}_{Fe_2O_3} ]$$
$$\because \Delta H^{\circ}$$ of free elements is zero, $$i.e., \Delta H^{\circ}_{Fe}=0; \:\Delta H^{\circ}_{Al}=0$$
$$= 2\times 0+ (-399.1)-[2\times 0+ (-196.5)]$$
$$\therefore \Delta H^{\circ}= -202.6 \: kcal$$
Question 9
1 / -0

The reaction that proceeds in the forward direction is:

A
$$SnC{l}_{4} + H{g}_{2}C{l}_{2} \longrightarrow SnC{l}_{2} + 2HgC{l}_{2}$$

B
$$N{H}_{4}Cl + NaOH \longrightarrow {H}_{2}O + N{H}_{3} + NaCl$$

C
$${Mn}^{2+} + 2{H}_{2}O + C{l}_{2} \longrightarrow Mn{O}_{2} + 4{H}^{+} + 2C{l}^{-}$$

D
$${S}_{4}{O}_{6}^{2-} + 2{I}^{-} \longrightarrow 2{S}_{2}{O}_{3}^{2-} + {I}_{2}$$

Solution

For a reaction to proceed in forward direction, it has to be spontaneous.
A spontaneous reaction is characterized by increase in entropy.

From the given reactions only option B shows an increase in entropy, because the number of molecules in the product is higher than that of the reactants.
Question 10
1 / -0

From the following data of $$\Delta H$$, of the following reactions:

$$C(s)+\frac{1}{2}O_2(g)\longrightarrow CO(g); \:                  \Delta H= -110 \, KJ$$
$$C(s)+ H_2O(g)\longrightarrow CO(g)+H_2(g); \: \Delta H= 132 \, KJ$$

The mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, keeping the temperature constant is:

A
$$\displaystyle \frac{0.8}{1}$$

B
$$\displaystyle \frac{0.6}{1}$$

C
$$\displaystyle \frac{0.4}{1}$$

D
$$\displaystyle \frac{0.2}{1}$$

Solution

The first reaction is exothermic and the second one is endothermic. If a mixture of steam and $$O_2$$ is passed over coke and temperature is kept constant, the conversion of each to CO should not show any heat change,

$$i.e., \: total \: heat \: evolved \: in \: I = total \: heat \: absorbed \: in \: II$$.

                      $$n_1 \times 2\times 110 = n_2\times 132$$

$$\therefore                        \displaystyle \frac{n_1}{n_2}=\frac{132}{110\times 2}=\frac{0.6}{1}$$

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Thermodynamics Test - 49

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Thermodynamics Test - 49

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Question 1

601.7

744.4

1346.1

147.7

Solution

Question 2

+640 kJ

+1280 kJ

-410 kJ

+410 kJ

Question 3

$$653.5\ kJ$$

$$-686.2\ kJ$$

$$-747.4\ kJ$$

None of these

Solution

Question 4

$$214.8$$

$$-195.0$$

$$-200.3$$

$$-256.45$$

Question 5

Lattice energy $$-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$$

Lattice energy $$-U\, =\, S\, -\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$$

Lattice energy $$-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, +\, E\, -\,Q$$

Lattice energy $$-U\, =\, S\, -\, I\, -\,\displaystyle \frac{D}{2}\, +\, E\, +\,Q$$

Solution

Question 6

$$1231.5J$$

$$1281.5J$$

$$781.5J$$

$$0$$

Question 7

9117

9441

9114

9141

Solution

Question 8

$$ -252.4 \: kcal$$

$$-135.5 \: kcal$$

$$ -202.6 \: kcal$$

none of the above

Solution

Question 9

$$SnC{l}_{4} + H{g}_{2}C{l}_{2} \longrightarrow SnC{l}_{2} + 2HgC{l}_{2}$$

$$N{H}_{4}Cl + NaOH \longrightarrow {H}_{2}O + N{H}_{3} + NaCl$$

$${Mn}^{2+} + 2{H}_{2}O + C{l}_{2} \longrightarrow Mn{O}_{2} + 4{H}^{+} + 2C{l}^{-}$$

$${S}_{4}{O}_{6}^{2-} + 2{I}^{-} \longrightarrow 2{S}_{2}{O}_{3}^{2-} + {I}_{2}$$

Solution

Question 10

$$\displaystyle \frac{0.8}{1}$$

$$\displaystyle \frac{0.6}{1}$$

$$\displaystyle \frac{0.4}{1}$$

$$\displaystyle \frac{0.2}{1}$$

Solution

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