Thermodynamics Test

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Thermodynamics Test - 59

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Question 1
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What are the most favourable conditions for the reaction;
$${ SO }_{ 2 }(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\leftrightharpoons { SO }_{ 3 }(g);\Delta { H }^{ o }=-ve$$ to occur?

A
Low temp and high press

B
Low temp and low press

C
High temp and low press

D
High temp and high press

Solution

Negative value of enthalpy change suggests exothermic reaction. Such reactions are favored at low temperature as heat released during the reaction raised the temperature and nullifies the effect of low temperature.
$$ \displaystyle \Delta n = \text { number of moles of gaseous products } - \text { number of moles of gaseous reactants } $$
$$ \displaystyle \Delta n = 1 - (1+ \dfrac {1 }{ 2 } )$$
$$ \displaystyle \Delta n =-\dfrac {1 }{ 2 } $$
Since$$ \displaystyle \Delta n < 0$$, the reaction is favored at high pressure. Increasing the pressure will shift the reaction towards forward direction (one with less number of moles of gaseous species). This will nullify the effect of increase in pressure.
Question 2
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Given,
$$C(s)+{O}_{2}(g)\rightarrow {CO}_{2}(g); \Delta H=-395\ kJ$$
$$S(s)+{O}_{2}(g)\rightarrow {SO}_{2}(g);\Delta H=-295\ kJ$$
$$C{S}_{2}(l)+3{O}_{2}(g)\rightarrow {CO}_{2}(g)+2{SO}_{2}(g);\Delta H=-1110\ kJ$$

The heat of formation of $$C{S}_{2}(l)$$ is:

A
$$250\ kJ$$

B
$$62.5\ kJ$$

C
$$31.25\ kJ$$

D
$$125\ kJ$$

Solution

Since, the formation reaction of $$CS_2$$ is,

$$C(s) + 2S(s) \rightarrow CS_2(l)$$

Since, the formation reaction of $$CS_2$$ can also be derived using the given reaction as,

$$1^{st} \text{reaction} +2\times 2^{nd} \text{ reaction} - 3^{rd} \text{ reaction}$$

Hence, the enthalpy of the reaction,

$$[1^{st} \text{reaction} +2\times 2^{nd} \text{ reaction} - 3^{rd} \text{ reaction}]$$ $$=\Delta H_{1^{st} \text{reaction}} +2\times \Delta H_{2^{nd} \text{ reaction}} -\Delta H_{3^{rd} \text{ reaction}}$$ $$=\text{heat of formation of CS}_2(l)$$

Hence, $$ \Delta H_{f(CS_2)} = -395 + 2\times (-295) - (-1110) = 125kJ$$

Hence, the answer is option $$D$$.
Question 3
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If $$\Delta H_{f}(H_{2}O)$$ is $$-285.20\ kJ\ mol^{-1}$$, then $$\Delta H_{f}^{\circ}(OH^{-})$$ is:

A
$$-227.90\ kJ\ mol^{-1}$$

B
$$+228.88\ kJ\ mol^{-1}$$

C
$$-343.52\ kJ\ mol^{-1}$$

D
$$+343.52\ kJ\ mol^{-1}$$

Solution

$$H^{+} + OH^{-}\rightarrow H_{2}O, \Delta H = -57.3\ kJ\ mol^{-1}$$ (standard value)

$$\Delta H = \Delta H_{f}^{\circ} (H_{2}O) - \Delta H_{f}^{\circ} (H^{+}) - \Delta H_{f}^{\circ}(OH^{-})$$

$$-57.3 = -285.2 - 0 - \Delta H_{f}^{\circ}(OH^{-})$$

$$\Delta H_{f}^{\circ} (OH^{-1}) = -227.9\ kJ\ mol^{-1}$$.
Question 4
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The enthalpy of combustion for the $$H_{2}$$, cyclohexene and cyclohexane are $$-241, -3800$$ and $$-3920kJ\ mol^{-1}$$, respectively. Heat of hydrogenation of cyclohexene is:

A
$$+121 kJ\ mol^{-1}$$

B
$$-121kJ\ mol^{-1}$$

C
$$+242kJ\ mol^{-1}$$

D
$$-242kJ\ mol^{-1}$$

Solution

Different combustion reactions are shown below.
$$\displaystyle C_6H_{12}+9O_2 \rightarrow 6CO_2+6H_2O $$

$$\displaystyle \Delta H_1 = -3920$$ kJ/mol   ......(1)

$$\displaystyle C_6H_{10}+8.5O_2 \rightarrow 6CO_2+5H_2O $$

$$\displaystyle \Delta H_2 = -3800$$ kJ/mol   ......(2)

$$\displaystyle H_{2}+O_2 \rightarrow H_2O $$ $$\displaystyle \Delta H_3 = -241$$ kJ/mol   ......(3)

The balanced chemical equation for the hydrogenation of cyclohexene is

Cyclohexene $$\displaystyle + H_2 \rightarrow $$ Cyclohexane   .....(4)

Add reaction (3) to reaction (2). Then substract reaction (1) to obtain reaction (4).

$$\displaystyle \Delta H_{4} = \Delta H_2- \Delta H_1+ \Delta H_3$$
$$\displaystyle \Delta H_4=-3800-(-3920)+-241$$
$$\displaystyle \Delta H_4=-121$$ kJ/mol

Hence, the heat of hydrogenation of cyclohexene is $$\displaystyle \Delta H=-121$$ kJ/mol
Question 5
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Standard heats of formation for $$C{Cl}_{4},{H}_{2}O,{CO}_{2}$$ and $$HCl$$ at $$298K$$ are $$-25.5,-57.8,-94.1$$ and $$-22.1kJ/mol$$ respectively.
For the reaction, what will be $$\Delta H$$?
$$C{Cl}_{4}+2{H}_{2}O\rightarrow {CO}_{2}+4HCl$$

A
$$36.4\ kJ$$

B
$$20.7\ kJ$$

C
$$-20.7\ kJ$$

D
$$-41.4\ kJ$$

Solution

For the given reaction,
$$CCl_4 +2H_2O \rightarrow CO_2 + 4HCl$$
$$\Delta H = \Delta H_{f(CO_2)} + $$ $$4\times$$ $$\Delta H_{f(HCl)}$$ - $$\Delta H_{f(CCl_4)}$$ - $$2\times$$ $$\Delta H_{f(H_2O)}$$
=> $$\Delta H = -94.1 + 4\times (-22.1) - (-25.5) - 2\times (-57.8) = -41.4 KJ$$
Hence, answer is option $$D$$.
Question 6
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Water is brought to boil under the pressure of $$1.0\ atm$$. When an electric current of $$0.50\ A$$ from a $$12\ V$$ supply is passed for $$300\ s$$ through resistance in thermal contact with it is found that $$0.798\ g$$ of water is vapourised. Calculate the molar internal energy change at boiling point $$(373.15\ K)$$.

A
$$37.5kJ\ { mol }^{ -1 }$$

B
$$3.75kJ\ { mol }^{ -1 }$$

C
$$42.6kJ\ { mol }^{ -1 }$$

D
$$4.26kJ\ { mol }^{ -1 }$$

Solution

$$\Delta H=$$ work done$$=i\times V\times t$$
$$=0.5\times 12\times 300=1.8\ kJ$$

Molar enthalpy of vaporisation
$$\Delta { H }_{ m }=\cfrac { \Delta H }{ moles\quad of\quad { H }_{ 2 }O\quad } =\cfrac { \Delta H }{ { n }_{ { H }_{ 2 }O } } $$

$$=\cfrac { 1.8kJ }{ \cfrac { 0.798 }{ 18 } } =40.6\ kJ{ mol }^{ -1 }$$

$$\Delta { H }_{ m }=\Delta { E }_{ m }+p\Delta V$$

$$\Delta { H }_{ m }=\Delta { E }_{ m }+\Delta { n }_{ g }RT\quad $$

$$\Delta { H }_{ m }=\Delta { E }_{ m }+RT\quad $$

[$$\therefore \Delta { n }_{ g }=1\quad for\quad { H }_{ 2 }O(l)\rightleftharpoons { H }_{ 2 }O(g)$$]

$$\therefore$$ Molar internal energy change,

$$\Delta { E }_{ m }=\Delta { H }_{ m }-RT$$

$$\Delta { E }_{ m }=40.6-8.314\times { 10 }^{ -3 }\times 373.15=37.5\ kJ { mol }^{ -1 }$$
Question 7
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Given: $$S+\dfrac { 3 }{ 2 } { O }_{ 2 }\longrightarrow S{ O }_{ 3 }+2x$$ $$kcal$$
$$S{ O }_{ 2 }+\dfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow S{ O }_{ 3 }+y$$ $$kcal$$

With the help of the above reactions, find out the heat of formation of $$S{ O }_{ 2 }$$.

A
$$\left( 2x-y \right) $$

B
$$\left( x+y \right) $$

C
$$\left( 2x+y \right) $$

D
$$\left( { 2x }/{ y } \right) $$

Solution

$$S+\dfrac { 3 }{ 2 } { O }_{ 2 }\longrightarrow S{ O }_{ 3 }+2x \: kcal$$.....(1)

$$S{ O }_{ 2 }+\dfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow S{ O }_{ 3 }+y \: kcal$$......(2)

The heat of formation of $$S{ O }_{ 2 }$$ can be represented as $$S+\ { O }_{ 2

}\longrightarrow S{ O }_{2 } ; \: \ \ \ \ \ \Delta H_f(SO_2)$$......(3)

Subtract equation (2) from equation (1) to obtain equation (3), we get-

$$ \displaystyle \Delta H_f(SO_2)=\left( 2x-y \right) $$ kcal

Hence, option A is correct.
Question 8
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A particular reaction at $${27}^{o}C$$ for which $$\Delta H > 0$$ and $$\Delta S> 0$$ is found to be non-spontaneous. The reaction may proceed spontaneously if:

A
the temperature is decreased

B
the temperature is kept constant

C
the temperature is increased

D
it is carried in open vessel at $${27}^{o}C$$

Solution

Gibb's energy equation: $$\Delta G=\Delta H-T\Delta S$$
For a spontaneous reaction, $$\Delta G$$ should be less than zero.

Therefore, if $$\Delta H>0$$ and $$\Delta S>0$$, the $$T\Delta S$$ term should be higher than $$\Delta H$$ for a spontaneous reaction. So, the temperature should be increased to make the reaction spontaneous.
Question 9
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Although the dissolution of ammonium chloride in water is an endothermic reaction, even then it is spontaneous because:

A
$$\Delta H$$ is positive, $$\Delta S$$ is -ve

B
$$\Delta H$$ is +ve, $$\Delta S$$ is zero

C
$$\Delta H$$ is positive, $$T\Delta S< \Delta H$$

D
$$\Delta H$$ is +ve, $$\Delta S$$ is positive and $$\Delta H< T\Delta S$$

Solution

Gibb's free energy equation: $$\Delta G=\Delta H-T\Delta S$$

For a spontaneous reaction, $$\Delta G<0$$
For endothermic reaction, $$\Delta H<0$$

For dissolution of ammonium chloride in water, $$\Delta S>0$$. The magnitude of $$T\Delta S$$ should be greater than that of $$\Delta H$$ so that $$\Delta G<0$$.

Therefore, the dissociation of ammonium chloride in water is an endothermic reaction but is still spontaneous because $$\Delta H$$ is $$+ve$$, $$\Delta S$$ is $$+ve$$ and $$\Delta H<T\Delta S$$.
Question 10
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The enthalpy of hydrogenation of benzene is $$-49.8\ kcal/mol$$ while its resonance energy is $$36.0\ kcal/mol$$. The enthalpy of formation of benzene is:

A
$$-4.6\ kcal$$

B
$$-28.6\ kcal/mol$$

C
$$-85.8\ kcal/mol$$

D
$$-13.8\ kcal/mol$$

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Thermodynamics Test - 59

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Question 1

Low temp and high press

Low temp and low press

High temp and low press

High temp and high press

Solution

Question 2

$$250\ kJ$$

$$62.5\ kJ$$

$$31.25\ kJ$$

$$125\ kJ$$

Solution

Question 3

$$-227.90\ kJ\ mol^{-1}$$

$$+228.88\ kJ\ mol^{-1}$$

$$-343.52\ kJ\ mol^{-1}$$

$$+343.52\ kJ\ mol^{-1}$$

Solution

Question 4

$$+121 kJ\ mol^{-1}$$

$$-121kJ\ mol^{-1}$$

$$+242kJ\ mol^{-1}$$

$$-242kJ\ mol^{-1}$$

Solution

Question 5

$$36.4\ kJ$$

$$20.7\ kJ$$

$$-20.7\ kJ$$

$$-41.4\ kJ$$

Solution

Question 6

$$37.5kJ\ { mol }^{ -1 }$$

$$3.75kJ\ { mol }^{ -1 }$$

$$42.6kJ\ { mol }^{ -1 }$$

$$4.26kJ\ { mol }^{ -1 }$$

Solution

Question 7

$$\left( 2x-y \right) $$

$$\left( x+y \right) $$

$$\left( 2x+y \right) $$

$$\left( { 2x }/{ y } \right) $$

Solution

Question 8

the temperature is decreased

the temperature is kept constant

the temperature is increased

it is carried in open vessel at $${27}^{o}C$$

Solution

Question 9

$$\Delta H$$ is positive, $$\Delta S$$ is -ve

$$\Delta H$$ is +ve, $$\Delta S$$ is zero

$$\Delta H$$ is positive, $$T\Delta S< \Delta H$$

$$\Delta H$$ is +ve, $$\Delta S$$ is positive and $$\Delta H< T\Delta S$$

Solution

Question 10

$$-4.6\ kcal$$

$$-28.6\ kcal/mol$$

$$-85.8\ kcal/mol$$

$$-13.8\ kcal/mol$$

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