Thermodynamics Test

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Thermodynamics Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

A reaction occurs spontaneously if:

A
T$$\Delta $$S > $$\Delta $$H and $$\Delta $$H is +ve and $$\Delta $$S is -ve

B
T$$\Delta $$S = $$\Delta $$H and both $$\Delta $$H and $$\Delta $$S are +ve

C
T$$\Delta $$S < $$\Delta $$H and both $$\Delta $$H and $$\Delta $$S are +ve

D
T$$\Delta $$S > $$\Delta $$H and both $$\Delta $$H and $$\Delta $$S are +ve

Solution

Hint: $$\Delta G <0$$ is a necessary and sufficient condition.

Step 1: Write the formula for $$\Delta G$$

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

Step 2: Equate $$\Delta G<0$$ as the condition for spontaneity.

$$\Delta H-T\Delta S < 0$$
$$\implies T\Delta S > \Delta H$$

But if $$\Delta H \rightarrow +ve, \Delta S \rightarrow -ve, $$ $$T\Delta S>\Delta H$$ Is not possible since the right hand side is positive which cannot be less than the negative left hand side. So we discard option $$A$$.

Final answer: Option $$D$$
Question 2
1 / -0

The heat of combustion for $$C$$, $$H_{2}$$ and $$CH_{4}$$ are $$-349$$ kJ/mol, $$-241.8$$ kJ/mol and $$-906.7$$ kJ/mol respectively. The heat of formation in kJ/mol of $$CH_{4}$$ is :

A
$$174.1$$

B
$$274.1$$

C
$$374.1$$

D
$$74.1$$

Solution

Given that,

$$C+O_{2}\rightarrow CO_{2}(g)\ ; \ \Delta H=-349$$ kJ/mol ..........$$(i)$$

$$H_{2}+\frac{1}{2}O_{2}\rightarrow H_{2}O(l)\ ;\ \Delta H=-241.8$$ kJ/mol ..........$$(ii)$$

$$CH_{4}+ 2O_{2}\rightarrow CO{_{2}}{(g)}+2H_{2}O(l)\ ; \ \Delta H=-906.7$$ kJ/mol .........$$(iii)$$

Now, we perform $$(i) + 2 \times (ii) - (iii)$$, we get

$$C(s)+2H_{2}(g)\rightarrow CH_{4}(g)$$

Therefore, $$\Delta H=-349+2(-241.8)+906.7$$ $$=74.1$$ kJ/mol

Hence, the heat of formation of $$CH_4$$ is $$74.1$$ kJ/mol.
Question 3
1 / -0

The change in free energy accompanied by the isothermal reversible expansion of 1 mole of an ideal gas when it doubles its volume is $$\Delta G_{1}$$. The change in free energy accompanied by a sudden isothermal irreversible doubling volume of 1 mole of the same gas is $$\Delta G_{2}$$. The ratio of $$\Delta G_{1}$$ and $$\Delta G_{2}$$ is :

A
$$1$$

B
$$\dfrac{1}{2}$$

C
$$2$$

D
$$\dfrac{3}{2}$$

Solution

The Gibb's free energy of the system is a state function.

Moving from state $$1$$ to $$2, \: \Delta G$$ will be the same irrespective of the type of process.

$$\therefore \dfrac{\Delta G_{1}}{\Delta G_{2}}=1$$
Question 4
1 / -0

Given that
$$Zn + \frac{1}{2} O_2 \rightarrow ZnO\quad\Delta{H}= + 84000 cal$$ ........... 1
$$Hg+\frac{1}{2} O_2 \rightarrow HgO\quad \Delta{H} =+ 21700 cal$$ ........... 2

The heat of reaction $$(\Delta H)$$ for,

$$Zn + HgO \rightarrow ZnO + Hg$$ is

A
105700 cal

B
62300 cal

C
-105700 cal

D
-62300 cal

Solution

The thermochemical reactions are as given below.

$$Zn + \frac{1}{2} O_2 \rightarrow ZnO + 84000 cal$$ ......1

$$Hg+\frac{1}{2} O_2 \rightarrow HgO + 21700 cal$$ ......2

The reaction 2 is reversed and added to the reaction 1 to obtain the reaction $$Zn + HgO \rightarrow ZnO + Hg$$

The heat of this reaction is $$84000-21700=62300$$
Question 5
1 / -0

Consider the following data: $$\Delta_fH^o(N_2H_4, l)=50kJ/mol, \Delta_fH^o(NH_3, g)=-46 kJ/mol$$,
$$B.E. (N-H)=393 kJ/mol$$ and $$B.E. (H-H)=436 kJ/mol$$,
$$\Delta_{vap}H(N_2H_4, l)=18 kJ/mol$$
Calculate the $$N-N$$ bond energy in kJ/mol for $$N_2H_4$$.

A
$$190 \ kJ/mol$$

B
$$-190 \ kJ/mol$$

C
$$95 \ kJ/mol$$

D
$$-95 \ kJ/mol$$

Solution
Question 6
1 / -0

The values of $$\Delta H$$ and $$\Delta S$$ for the reaction, C(graphite) + $$CO_2(g)\rightarrow 2CO(g)$$ are 170 kJ and $$170\; JK^{-1}$$ respectively. This reaction will be spontaneous at:

A
510 K

B
710 K

C
910 K

D
1010 K

Solution

For a spontaneous reaction, $$\triangle G=$$ Negative
And at equilibrium $$\triangle G=0$$ i.e $$\triangle H=T\triangle S$$
$$\Rightarrow T=\cfrac {170\times 10^{3}}{170}$$
$$\Rightarrow T=1000K$$
Therefore, at this temperature, reaction is at equilibrium i.e $$\triangle G=0$$ and above this temperature $$\triangle G$$ will be negative and reaction will be spontaneous.
Question 7
1 / -0

A 1000 gm sample of water is reacted with an equimolar amount of CaO (both at same initial temperature of $$25^{\circ}C$$). Assuming the container to be adiabatic, the final temperature of the product is approx :

Given : $$CaO + H_2O\rightarrow Ca(OH)_2,\ \Delta H=-65.2 KJ/mol $$
Specific heat of $$Ca(OH)_2 = 1.2 \: J/gm^{\circ}C$$

A
$$735^{\circ}C$$

B
$$760^{\circ}C$$

C
$$746^{\circ}C$$

D
$$789^{\circ}C$$

Solution

$$\text{Total heat released}=65.2\times 10^3\times \frac{1000}{18}$$

$$\text{Mass of }Ca(OH)_2 \text{produced} = \frac{1000}{18}\times74 \: gm$$

Now, by energy balance equation:

$$\frac{1000}{18}\times74 \times 1.2(T_f-25)=\frac{1000}{18}\times 65.2\times 10^3$$

$$T_f=759.23^0C\approx 760^0C$$
Question 8
1 / -0

If $$\Delta H{_{f}}^{0}$$ for $$Ag^{+} (\infty$$ diluted), $$NO{_{3}}^{-} (\infty$$ diluted), $$Cl^{-} (\infty$$ diluted) and $$AgCl_{(S)}$$ are $$-105.579, -207.36, -167.159$$ and $$-127.068$$ respectively. Calculate the enthalpy change for the reaction $$AgNO_{3(aq)}+HCl_{(aq)}\rightarrow AgCl_{(S)}+HNO_{3} (aq)$$.

A
21.471 kJ/mol

B
145.67 kJ/mol

C
65.488 kJ/mol

D
None

Solution

$$\Delta H^{\circ}_{reaction}=[\Delta H_{f}(AgCl)+\Delta H_{f}(H^{+})+\Delta H_{f}(NO{_{3}}^{-})]$$

$$- [ \Delta H^{\circ}_{f}(Ag^{+}) +\Delta H^{\circ}_{f}(NO{_{3}}^{-})+\Delta H^{\circ}_{f}(Cl^{-})+\Delta H^{\circ}_{f}(H^{+})]$$

$$\Delta H^{\circ}_{reaction}=-127.068-\left [- 105.579-167.159 \right ]$$

$$\Delta H^{\circ}_{reaction}=145.67$$ kJ/mol.
Question 9
1 / -0

Following are the thermochemical reactions:
$$C$$ (graphite) $$+ \dfrac{1}{2}O_2 \rightarrow CO \ ; \Delta H = - 110.5$$ kJ/mol
$$CO + \dfrac{1}{2} O_2 \rightarrow CO_2 \ ; \Delta H = - 283.2$$ kJ/mol
The heat of reaction (in kJ/mol) for the following reaction is:
$$C (\text{graphite}) + O_2 \rightarrow CO_2$$

A
$$+393.7$$

B
$$-393.7$$

C
$$-172.7$$

D
$$+172.7$$

Solution

The thermochemical reactions are as given below:
$$C$$ (graphite) $$+ \dfrac{1}{2}O_2 \rightarrow CO ; \Delta H = - 110.5 $$ kJ/mol $$...(1)$$
$$CO + \dfrac{1}{2} O_2 \rightarrow CO_2 ; \Delta H = - 283.2$$ kJ/mol $$......(2)$$
The above two reactions are added to obtain the following reaction:
$$C$$ (graphite) $$+ O_2 \rightarrow CO_2$$
By applying Hess's law, the enthalpy change will be $$- 110.5+(- 283.2)=-393.7$$ kJ/mol
Hence, the heat of the given reaction is $$-393.7$$ kJ/mol.
Question 10
1 / -0

Find the heat change in the reaction
$$NH_3(g) + HCl (g) \rightarrow NH_4 Cl(s)$$ from the following data
$$NH_3 (g) + aq \rightarrow NH_3 (aq),$$ $$\Delta H = - 8.4 K. Cal.$$
$$HCl(g) + aq \rightarrow HCl(aq),$$   $$\Delta H = -17.3 K. Cal.$$
$$NH_3(aq) + HCl(aq) \rightarrow NH_4 Cl(aq), \Delta H = - 12.5 K. Cals.$$
$$NH_4 Cl(s) + aq \rightarrow NH_4 Cl(aq),    \Delta H = + 3.9 K. Cal.$$

A
-42.1

B
-23.3

C
+34.3

D
+42.1

Solution

The thermochemical reactions are as given below.
$$NH_3 (g) + aq \rightarrow NH_3 (aq),\Delta H = - 8.4 K. Cal.$$
$$HCl(g) + aq \rightarrow HCl(aq),\Delta H = -17.3 K. Cal.$$
$$NH_3(aq) + HCl(aq) \rightarrow NH_4 Cl(aq), \Delta H = - 12.5 K. Cal.$$
$$NH_4 Cl(s) + aq \rightarrow NH_4 Cl(aq),                \Delta H = + 3.9 K. Cal.$$
The reaction (4) is reversed and added to the reactions (1), (2) and (3) to obtain the reaction $$NH_3(g) + HCl (g)   \rightarrow NH_4 Cl(s)$$.
Hence, the heat change of the reaction is $$-8.4-17.3-12.5-3.9=-42.1$$K. Cal

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Thermodynamics Test - 47

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Question 1

T$$\Delta $$S > $$\Delta $$H and $$\Delta $$H is +ve and $$\Delta $$S is -ve

T$$\Delta $$S = $$\Delta $$H and both $$\Delta $$H and $$\Delta $$S are +ve

T$$\Delta $$S < $$\Delta $$H and both $$\Delta $$H and $$\Delta $$S are +ve

T$$\Delta $$S > $$\Delta $$H and both $$\Delta $$H and $$\Delta $$S are +ve

Solution

Question 2

$$174.1$$

$$274.1$$

$$374.1$$

$$74.1$$

Solution

Question 3

$$1$$

$$\dfrac{1}{2}$$

$$2$$

$$\dfrac{3}{2}$$

Solution

Question 4

105700 cal

62300 cal

-105700 cal

-62300 cal

Solution

Question 5

$$190 \ kJ/mol$$

$$-190 \ kJ/mol$$

$$95 \ kJ/mol$$

$$-95 \ kJ/mol$$

Solution

Question 6

510 K

710 K

910 K

1010 K

Solution

Question 7

$$735^{\circ}C$$

$$760^{\circ}C$$

$$746^{\circ}C$$

$$789^{\circ}C$$

Solution

Question 8

21.471 kJ/mol

145.67 kJ/mol

65.488 kJ/mol

None

Solution

Question 9

$$+393.7$$

$$-393.7$$

$$-172.7$$

$$+172.7$$

Solution

Question 10

-42.1

-23.3

+34.3

+42.1

Solution

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