Thermodynamics Test

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Thermodynamics Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

A heat engine operating between 227 deg C and 27 deg C absorbs 1 kcal of heat from the 227 deg C reservoir per cycle. Calculate
(1) the amount of heat discharged into the low temperature reservoir.
(2) the amount of work done per cycle.
(3) the efficiency of cycle.

A
0.4 kcal, 0.6 kcal, 40%

B
0.6 kcal, 0.4 kcal, 40%

C
0.4 kcal, 0.6 kcal, 60%

D
0.7 kcal, 0.4 kcal, 40%

Solution

Calculation of work done:
$$\displaystyle T' = 227+273=500$$ K
$$\displaystyle T=27+273=300$$ K
$$\displaystyle q'=1$$ 1 kcal
$$\displaystyle \dfrac {w}{q'} = \dfrac {T'-T}{T'}$$
$$\displaystyle w=q(\dfrac {T'-T}{T'}) $$
Putting the various values in the above reaction
$$\displaystyle w=1(\dfrac {500-300}{500})=1 \times \dfrac {200}{500}=0.4$$ kcal.

Calculation of heat discharged q
$$\displaystyle q'-q=w$$
$$\displaystyle 1-q=0.4$$
$$\displaystyle q=0.6$$ kcal

Calculation of efficiency
$$\displaystyle \eta = \dfrac {w}{q'}=0.4$$ kcal or 40%
Question 2
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Calculate enthalpy for formation of ethylene from the following data:

(I) $$C(graphite) + O_2 (g) \rightarrow CO_2 (g); \ \ \ \Delta H = -393.5 kJ$$
(II) $$H_2(g) + \dfrac{1}{2} O_2 (g) \rightarrow H_2O(l); \ \ \ \ \Delta H = - 286.2 kJ$$
(III) $$C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2 O(l); \ \ \ \ \Delta H = - 1410.8 kJ$$

A
54.1 kJ

B
44.8 kJ

C
51.4 kJ

D
48.4 kJ

Solution

(I) $$C(graphite) + O_2 (g) \rightarrow CO_2 (g) $$ ---- 1

(II) $$H_2(g) + \dfrac{1}{2} O_2 (g) \rightarrow H_2O(l)$$ ----- 2

(III) $$C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2 O(l)$$ ---- 3

Desired reaction:
$$2C(grapite)+2H_2(g)\rightarrow C_2H_4(g)$$

The desired above reaction can be obtained by multiplying eqn. 1 and 2 by 2 and adding them followed by subtracting eqn. 3 from the sum.

Enthalpy of formation can be calculated as:

$$\Delta H_f=(2\times Reaction\ I+2\times Reaction\ II-Reaction\ III)$$

$$\Delta H_f=(-2\times393.5-2\times286.2+1410.8)=51.4\ kJ$$
Question 3
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What will be the heat formation of methane; if the heat of combustion of carbon is '$$-x$$' $$kJ$$, heat of formation of water is '$$-y$$' $$kJ$$ and heat of combustion of methane is '$$-z$$' $$kJ$$?

A
$$(-x-y-z)kJ$$

B
$$(-z-x+2y)kJ$$

C
$$(-x-2y-z)kJ$$

D
$$(-x-2y+z)kJ$$

Solution

Formation of methane $$\Rightarrow \quad C+2{ H }_{ 2 }O\rightarrow { CH }_{ 4 }+{ O }_{ 2 }$$
$$\therefore $$ Heat of formation of methane $$={ \Delta }_{ c }\left( C \right) -{ \Delta }_{ c }\left( { CH }_{ 4 } \right) -2\times { \Delta }_{ f }\left( { H }_{ 2 }O \right) $$
$$=-x-\left( -z \right) -2y$$
$$=\left( -x+z-2y \right) kJ$$
$$=(-x-2y+z)kJ$$
Question 4
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Which one of the following is not applicable for a thermochemical equation?

A
It tell about physical state of reactants and products

B
It tells whether the reaction is spontaneous

C
It tells whether the reaction is exothermic or endothermic

D
It tells about the allotropic form (if any) of the reactants

Solution

A thermochemical equation tells us about the physical state of reactants and products as it is shown in the reaction. It also tells us about the reaction whether it is endothermic or exothermic from the heat in reactant or product. It also tells us about the allotropic form if any but it doesn't tells us about the spontaneity of a reaction. Spontaniety of a reaction is the measures of its randomness and can be calculated using formulations.
Question 5
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Super cooled water is liquid water that has been cooled below its normal freezing point. This state is thermodynamically :

A
unstable and tends to freeze into ice spontaneously

B
stable and tends to freeze Into ice spontaneously

C
stable and tends to fuse into liquid spontaneously

D
unstable and tends to fuse into liquid spontaneously

Solution

Super cooled water is liquid water that has been cooled below its normal freezing point. This state is thermodynamically unstable and tends to freeze into ice spontaneously. Below normal freezing point, water cannot exist in liquid state. It has to exist in solid state. Due to this, the conversion liquid water $$\displaystyle \rightarrow $$ solid ice is spontaneous.
Question 6
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If $${H}_{2}+\cfrac{1}{2}{O}_{2}\rightarrow {H}_{2}O;\Delta H=-68.09kcal$$
$$K+{H}_{2}O+water \rightarrow KOH(aq.)+\cfrac{1}{2}{H}_{2};\Delta H=-48.0kcal$$
$$KOH+water\rightarrow KOH(aq); \Delta H=-14.0kcal$$
the heat of formation of $$KOH$$ is:

A
$$-68.39+48-14.0$$

B
$$-68.39-48.0+14.0$$

C
$$+68.39-48.0+14.0$$

D
$$+68.39+48.0-14.0$$

Solution

(i) $${ H }_{ 2 }+\dfrac { 1 }{ 2 } { O }_{ 2 }\rightarrow { H }_{ 2 }O\quad \quad \quad ;\quad { \Delta H }_{ f }=-68.09Kcal$$
(ii) $$K+{ H }_{ 2 }O+water\rightarrow KOH\left( aq \right) +\dfrac { 1 }{ 2 } { H }_{ 2 }\quad ;\quad { \Delta H }_{ f }=-48Kcal$$
(iii) $$KOH+water\rightarrow KOH\left( aq \right) \quad \quad ;\quad { \Delta H }_{ f }=-14.0Kcal$$
(i) + (ii) - (iii) gives heat of formation of $$KOH$$
$$\therefore $$ Heat of formation of $$KOH=\left( -68.09-48.0+14.0 \right) Kcal$$
Question 7
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From the following reactions at $$298\ K$$.

(A) $$CaC_{2}(s) + 2H_{2}O(l) \rightarrow Ca(OH)_{2}(s) + C_{2}H_{2} (g);\ \Delta H^{\circ}= - 127.9\ kJ\ mol^{-1}$$

(B) $$Ca(s) + \dfrac {1}{2} O_{2}(g) \rightarrow CaO(s) ;\ \Delta H=- 635.1kJ\ mol^{-1}$$

(C) $$CaO(s) + H_{2}O(l) \rightarrow Ca(OH)_{2}(s);\ \Delta H=- 65.2\ kJ\ mol^{-1}$$

(D) $$C(s) + O_{2}(s) \rightarrow CO_{2}(s) ;\ \Delta H=- 393.5\ kJ\ mol^{-1}$$

(E) $$C_{2}H_{2}(g) + \dfrac {5}{2}O_{2}(g) \rightarrow 2CO_{2}(g) + H_{2}O(l);\ \Delta H= - 1299.58\ kJ\ mol^{-1}$$

Calculate the heat of formation of $$CaC_{2}(s)$$ at $$298\ K$$.

A
$$-59.82\ kJ\ mol^{-1}$$

B
$$+59.82\ kJ\ mol^{-1}$$

C
$$-190.22\ kJ\ mol^{-1}$$

D
$$+190.22\ kJ\ mol^{-1}$$
Question 8
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$$S+{O}_{2}\rightarrow {SO}_{2}; \Delta H=-298.2kJ$$
$${ SO }_{ 2 }+\cfrac { 1 }{ 2 }{O}_{2} \rightarrow { SO_3 }; \Delta H=-98.7kJ$$
$${SO}_{3}+{H}_{2}O\rightarrow {H}_{2}{SO}_{4};\Delta H=-130.2kJ$$
$${H}_{2}+\cfrac { 1 }{ 2 } { O }_{ 2 }\rightarrow {H}_{2}O;\Delta H=-227.3kJ$$
The heat of formation of $${H}_{2}{SO}_{4}$$ will be:

A
$$-754.4kJ$$

B
$$+320.5kJ$$

C
$$-650.3kJ$$

D
$$-433.7kJ$$

Solution

Since, on adding all the 4 reaction given in question we will get a final reaction,
$$S + 2O_2 + H_2\rightarrow H_2SO_4$$
Which is basically the formation reaction of $$H_2SO_4$$
Hence, the heat of formation of $$H_2SO_4$$ = sum of heat of reactions of all four reaction, i.e,
$$\Delta H_{f(H_2SO_4)} = \Delta H_1 + \Delta H_2 + \Delta H_3 +\Delta H_4 = -298.2 -98.7 - 130.2 -227.3 = -754.4KJ.$$
hence, answer is option $$A$$.
Question 9
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$$\Delta {H}_{f(x)},\Delta {H}_{f(y)},\Delta {H}_{f(R)}$$ and $$\Delta {H}_{f(S)}$$ denote the enthalpies of formation of $$x,y,R$$ and $$S$$ respectively. The enthalpy of the reaction $$x+y\rightarrow R+S$$ is:

A
$$\Delta {H}_{f(x)}+\Delta {H}_{f(y)}$$

B
$$\Delta {H}_{f(R)}+\Delta {H}_{f(S)}$$

C
$$\Delta {H}_{f(x)}+\Delta {H}_{f(y)}-\Delta {H}_{f(R)}-\Delta {H}_{f(S)}$$

D
$$\Delta {H}_{f(R)}+\Delta {H}_{f(S)}-\Delta {H}_{f(x)}-\Delta {H}_{f(y)}$$

Solution

For any reaction the enthalpy change of the reaction is given by,
$$\Delta H = \text{sum of heat of formation of products in stoichiometric ratio - sum of heat of formation of reactants in stoichiometric ratio}$$
i.e, for any reaction
$$aA + bB \rightarrow cC + dD$$
$$\Delta H = c\Delta H_{f(C)} + d\Delta H_{f(D)} - a\Delta H_{f(A)} - b\Delta H_{f(B)}$$
Hence, for the given reaction, $$x+y\rightarrow R+S$$,
$$\Delta H = \Delta H_{f(R)} + \Delta H_{f(S)} - \Delta H_{f(x)} - \Delta H_{f(y)}$$
Hence, answer is option $$D$$.
Question 10
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Given that the bond energies of $$:N\equiv N$$ is $$946$$ kJ $$mol^{-1}$$ $$H-H$$ is $$435 $$ kJ $$mol^{-1}$$, $$N-N$$ is $$159$$ kJ $$mol^{-1}$$, and $$N-H$$ is $$389$$ kJ $$mol^{-1}$$, the heat of formation of hydrazine in the gas phase in kJ $$mol^{-1}$$ is:

A
$$833$$

B
$$101$$

C
$$334$$

D
$$1264$$

Solution

$$N_2+2H_2\rightarrow N_2H_4$$

$$\Delta H_f=B.E(reactants)-B.E(products)$$

$$\Delta H_f=946+2\times435-(4\times389+159)=101\ kJ/mol$$

Hence option $$B$$ is correct.

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Thermodynamics Test - 58

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Thermodynamics Test - 58

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Question 1

0.4 kcal, 0.6 kcal, 40%

0.6 kcal, 0.4 kcal, 40%

0.4 kcal, 0.6 kcal, 60%

0.7 kcal, 0.4 kcal, 40%

Solution

Question 2

54.1 kJ

44.8 kJ

51.4 kJ

48.4 kJ

Solution

Question 3

$$(-x-y-z)kJ$$

$$(-z-x+2y)kJ$$

$$(-x-2y-z)kJ$$

$$(-x-2y+z)kJ$$

Solution

Question 4

It tell about physical state of reactants and products

It tells whether the reaction is spontaneous

It tells whether the reaction is exothermic or endothermic

It tells about the allotropic form (if any) of the reactants

Solution

Question 5

unstable and tends to freeze into ice spontaneously

stable and tends to freeze Into ice spontaneously

stable and tends to fuse into liquid spontaneously

unstable and tends to fuse into liquid spontaneously

Solution

Question 6

$$-68.39+48-14.0$$

$$-68.39-48.0+14.0$$

$$+68.39-48.0+14.0$$

$$+68.39+48.0-14.0$$

Solution

Question 7

$$-59.82\ kJ\ mol^{-1}$$

$$+59.82\ kJ\ mol^{-1}$$

$$-190.22\ kJ\ mol^{-1}$$

$$+190.22\ kJ\ mol^{-1}$$

Question 8

$$-754.4kJ$$

$$+320.5kJ$$

$$-650.3kJ$$

$$-433.7kJ$$

Solution

Question 9

$$\Delta {H}_{f(x)}+\Delta {H}_{f(y)}$$

$$\Delta {H}_{f(R)}+\Delta {H}_{f(S)}$$

$$\Delta {H}_{f(x)}+\Delta {H}_{f(y)}-\Delta {H}_{f(R)}-\Delta {H}_{f(S)}$$

$$\Delta {H}_{f(R)}+\Delta {H}_{f(S)}-\Delta {H}_{f(x)}-\Delta {H}_{f(y)}$$

Solution

Question 10

$$833$$

$$101$$

$$334$$

$$1264$$

Solution

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