Thermodynamics Test

Result

Thermodynamics Test - 51

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If the resonance energy of $$NO_2(:O--N==O:)$$ is X kJ The measured enthalpy formation of $$NO_2(\Delta_f H^{ \ominus } )$$ is $$34 kJ mol^{ -1 }$$. The bond energies given are:
$$N--O \Rightarrow 222 kJ mol^{ -1 }$$
$$N\equiv N \Rightarrow 946 kJ mol^{ -1 }$$
$$O==O \Rightarrow 498 kJ mol^{ -1 }$$
$$N==O \Rightarrow 607 kJ mol^{ -1 }$$
Find out the value of X

A
$$X= -108 $$

B
$$X= +108 $$

C
$$X= -59$$

D
None of these

Solution

Sol. $$R.E. = \Delta_f H(observed) - \Delta_f H(calculated)$$
The balanced chemical reaction for the formation of nitrogen dioxide is as shown.
$$\frac { 1 }{ 2 } N_2(g) + O_2(g)\longrightarrow NO_2(g)$$ $$\Delta_f H_{ cal } = \text {(BE of reactants - BE of products)}$$ $$=\begin{pmatrix} \frac { 1 }{ 2 } \text {BE of } N_2 +\text {BE of } O_2\end{pmatrix} - \begin{pmatrix} \frac { 1 }{ 2 } \text {BE of } N==O + \text {BE of } N--O\end{pmatrix}$$
$$=\begin{pmatrix} \frac { 1 }{ 2 }\times 946 + 498\end{pmatrix} - (607 + 222)$$
$$= 971-829$$
$$=142 kJ$$ $$\Delta H_{(observed)}$$
$$= 34 kJ$$
$$\therefore $$ Resonacne .Energy $$ = 34-142$$
$$= -108 kJ$$
Hence, the value of $$X$$ is $$-108$$.
Question 2
1 / -0

Find the $$\Delta H$$ of the following reaction. $$OF_2(g) + H_2O(g)\longrightarrow O_2(g) + 2HF(g)$$. Average bond energies of $$O-F$$, $$O-H$$, $$O=O$$ and $$H-F$$ are $$44$$, $$111$$, $$118$$ and $$135 $$ kcal mol$$^{ -1 }$$, respectively.

A
$$\Delta H= -78\ kcal$$

B
$$\Delta H= +78\ kcal$$

C
$$\Delta H= -498\ kcal$$

D
$$\Delta H= +498 \ kcal$$

Solution

As we know,
$$\Delta H=$$ Bond energy data of formation of bond + Bond energy data of dissociation of bond

So for $$OF_2 (g)+H_2O(g)⟶O_2(g)+2HF(g)$$,

$$\Delta H_{reac} = \Delta H_{BE reactant}- \Delta H_{BEproduct}$$

$$\Delta H = 2\times 44+2\times 111 - 118 - 2\times 135 = -78\ kcal/mol$$

Option A is correct.
Question 3
1 / -0

The thermo-chemical equation for solid and liquid rocket fuel are given below:
$$2Al(s) + 1\frac { 1 }{ 2 }O_2(g)\longrightarrow Al_2O_3(s), \Delta H = -1667.8 kJ$$

$$H_2(g) +\frac { 1 }{ 2 }O_2(g)\longrightarrow H_2O(l), \Delta H = -285.9 kJ$$,

Then, $$\Delta H$$ for the reaction : $$Al_2O_3(s)\longrightarrow 2Al(s) + 1\frac { 1 }{ 2 }O_2(g)$$ is :

A
$$\ 1667.8\ kJ mol^{ -1 }$$

B
$$\ 1222.3\ kJ mol^{ -1 }$$

C
$$\ 1458.2\ kJ mol^{ -1 }$$

D
$$\ 1785.3\ kJ mol^{ -1 }$$

Solution

As given,
$$2Al(s) + 1\frac { 1 }{ 2 }O_2(g)\longrightarrow Al_2O_3(s), \Delta H = -1667.8 kJ$$

and by this we can see that, for reverse reaction :

$$Al_2O_3(s)\longrightarrow 2Al(s) + 1\frac { 1 }{ 2 }O_2(g); \Delta H = 1667.8 kJ$$
Question 4
1 / -0

Calculate the $$\Delta H^{ \ominus }$$ for the reduction of $$Fe_2O_3(s)$$ by $$Al(s)$$ at $$25^{ \circ }C$$. The enthalpies of formation of $$Fe_2O_3(s)$$ and $$Al_2O_3$$ are $$-825.5$$ and $$-1675.7 kJ mol^{ -1 }$$ respectively.

A
$$\Delta H= - 850.2 kJ mol^{ -1 }$$

B
$$\Delta H= +850.2 kJ mol^{ -1 }$$

C
$$\Delta H= - 2500 kJ mol^{ -1 }$$

D
None of these

Solution

The reduction of $$Fe_2O_3$$ by Al is represented by following chemical reaction:
$$ Fe_2O_3(s) + 2Al(s)\longrightarrow 2Fe(s) + Al_2O_3; \ \ \ \ \ \ \Delta H = ?$$ ......(1)

The balanced chemical reaction for the formation of $$Fe_2O_3$$ is as shown below.
$$3Fe(s) + \cfrac { 3 }{ 2 }O_2(g)\longrightarrow Fe_2O_3; \ \ \ \ \ \ \ \ \ \Delta H_1 = -825.5 kJ mol^{ -1 }$$ ......(ii)

The balanced chemical reaction for the formation of $$Al_2O_3$$ is as shown below.
$$2Al(s) +\cfrac { 3 }{ 2 }O_2(g)\longrightarrow Al_2O_3; \ \ \ \ \ \ \ \ \Delta H_2 = -1675.7 kJ mol^{ -1 }$$......(iii)

Subtract the enthalpy change for the third reaction from the enthalpy change for the second reaction to obtain the enthalpy change for the first reaction.

$$\Delta H = \Delta H_1 - \Delta H_2$$ $$= -1675.7 - (-825.5)$$ $$= - 850.2 kJ mol^{ -1 }$$

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

How much heat is liberated when one mole of gaseous $$Na^{ \oplus }$$ combines with one mole of $$Cl^{ \ominus }$$ ion to form solid $$NaCl$$.

Use the data given below:
$$Na(s) + \frac { 1 }{ 2 } Cl_2(g) \longrightarrow NaCl(s)$$; $$\Delta H = -98.23 kcal$$
$$Na(s)\longrightarrow Na(g)$$;$$\,\,\,\, \Delta H = +25.98 kcal$$
$$Na(g)\longrightarrow Na^{ \oplus } + e^ { - }$$;$$\,\,\,\,\,\,\,\Delta H = +120.0 kcal$$
$$Cl_2(g)\longrightarrow 2Cl(g)$$;$$\,\,\,\,\,\,\,\,\Delta H = +58.02 kcal$$
$$Cl^{ \ominus } (g)\longrightarrow Cl(g) + e^{ - }$$;$$\,\,\,\,\,\,\,\Delta H = +87.3 kcal$$

A
$$\Delta H = -185.92 Kcal$$

B
$$\Delta H = +185.92 Kcal$$

C
$$\Delta H = 0 Kcal$$

D
None of these

Solution

$$Na^{ \oplus } (g) + Cl^{ \ominus } (g)\longrightarrow NaCl(s)$$$$ ;\,\,\,\,\,\,\Delta H = ?$$

Given:

$$Na(s) + \frac { 1 }{ 2 } Cl_2(g)\longrightarrow NaCl(s)$$$$;\,\,\,\,\,\,\,\Delta H = -98.23 kcal$$

$$Na(s)\longrightarrow Na(g)$$ $$;\,\,\,\,\,\,\,\,\Delta H = 25.98 kcal$$

$$Na(g)\longrightarrow Na^{ \oplus } (g) + e^{ - }$$ $$;\,\,\,\,\,\,\,\,\Delta H = 120.0 kcal$$

$$\frac { 1 }{ 2 } Cl_2(g)\longrightarrow Cl(g)$$$$;\,\,\,\,\,\,\,\,\Delta H = 58.02\times \frac { 1 }{ 2 } kcal$$

$$Cl^{ \ominus} (g)\longrightarrow Cl(g) + e^{ - }$$$$;\,\,\,\,\,\,\,\,\,\Delta H =87.3 kcal$$

Rewriting equations:

$$Na(s) + \frac { 1 }{ 2 } Cl_2(g)\longrightarrow NaCl(s)$$$$;\,\,\,\,\,\,\,\,\Delta H_1 = -98.23$$

$$Na(g)\longrightarrow Na(s)$$ $$;\,\,\,\,\,\,\,\,\Delta H_2 = -25.98$$

$$Na^{ \oplus } (g) + e^{ - }\longrightarrow Na(g)$$ $$;\,\,\,\,\,\,\,\,\Delta H_3 = -120.0$$

$$\frac { 1 }{ 2 } [2Cl (g)\longrightarrow Cl_2 (g) ]$$$$;\,\,\,\,\,\,\,\,\Delta H_4 = -58.02\times \frac { 1 }{ 2 }$$

$$Cl^{ \ominus} (g)\longrightarrow Cl(g) + e^{ - }$$$$;\,\,\,\,\,\,\,\,\,\Delta H_5 = 87.3$$

$$Na^{ \oplus }(g) + Cl^{ \ominus }\longrightarrow NaCl(s)$$

$$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 + \Delta H_5$$
$$\,\,\,\,\,\,\,\,\ = -98.23 - 25.98 - 120.0 - \frac{ 1 }{ 2 }\times 58.02 + 87.3$$
$$\,\,\,\,\,\,\, \ = -185.92 Kcal$$
Question 6
1 / -0

The standard enthalpy of formation of $$FeO$$ and $$Fe_2O_3$$ is $$-65 \: kcal \: mol^{-1}$$ and $$-197 \: kcal \: mol^{-1}$$ respectively. If a mixture containing $$FeO$$ and $$FeO_3$$ in $$2 : 1$$ mole ratio on oxidation is changed into $$1 : 2$$ mole ratio, thermal energy in kcal released per mole of mixture will be :

A
$$13.4$$

B
$$26.8$$

C
$$30.2$$

D
none of these

Solution

$$\displaystyle Fe+\frac{1}{2}O_2\longrightarrow FeO; \:       \Delta H=-65 \: kcal \: mol^{-1}$$          . ..(i)
               $$\displaystyle 2Fe+\frac{3}{2}O_2\longrightarrow Fe_2O_3; \: \Delta H=-197 \: kcal \: mol^{-1}$$        ...(ii)
By eq, $$[(ii) - 2 \times (i)]$$
Also,    $$\displaystyle 2FeO+\frac{1}{2}O_2\longrightarrow Fe_2O_3; \: \Delta H=-67 \: kcal \: mol^{-1}$$          ...(iii)
Let $$a$$ mole of FeO and $$b$$ mole of $$Fe_2O_3$$, are present such that
               $$\displaystyle a+b=1 \: and \:\frac{a}{b}=2$$
                  $$\displaystyle a=\frac{2}{3} \:and \:b=\frac{1}{3}$$
                  $$2FeO+\frac{1}{2}O_2\longrightarrow Fe_2O_3$$
Initial                     $$a$$ $$b$$
After oxidation $$(a - 2a')$$      $$b+ a'$$
Given,      $$\displaystyle \frac{a}{b}=2 \:and \: \frac{a - 2a'}{b+ a'}=\frac{1}{2}$$
$$\displaystyle \therefore   \frac{a - 2a'}{\frac{a}{2}+ a'}=\frac{1}{2}$$
$$\displaystyle \therefore   2a-4a'=\frac{a}{2}+ a'$$
$$\displaystyle \therefore      a'=\frac{3a}{10}=\frac{3\times 2}{10\times 3}=\frac{1}{5}$$
$$\therefore FeO \: used = 2a' = 2/5 \: mol$$
$$\because 2 \: mole \: FeO \: gives \: heat = 67 \: kcal$$
$$\displaystyle \therefore \frac{2}{5} mole \: FeO \: gives \: heat =\frac{67\times 2}{2\times 5}= 13.4 \: kcal $$
Question 7
1 / -0

When $$10 g$$ of $$Al$$ is used for reduction in each of the following alumino thermic reactions, which reaction would generate more heat and by how much ?
a. $$2Al + Cr_2O_3\rightarrow Al_2O_3 + 2Cr$$
b. $$2Al + Fe_2O_3\rightarrow Al_2O_3 + 2Fe$$

Standard heat of formation of $$Al_2O_3$$, $$Cr_2O_3$$ and $$Fe_2O_3$$ are $$-1676 kJ$$, $$-1141 kJ$$ and $$-822.2 kJ$$, respectively.

A
Reaction (b) produces more heat by $$59.04 kJ$$

B
Reaction (a) produces more heat by $$59.04 kJ$$

C
Reaction(a) and (b) produces equal heat

D
None of these

Solution

Given,
$$2Al + \dfrac{ 3 }{ 2 }O_2\rightarrow Al_2O_3; \Delta H_1 = -1676 kJ......(i)$$

$$2Cr + \dfrac { 3 }{ 2 }O_2\rightarrow Cr_2O_3; \Delta H_2 = -1141 kJ....(ii)$$

$$2Fe + \dfrac { 3 }{ 2 }O_2\rightarrow Fe_2O_3; \Delta H_3 = -822.2 kJ ....(iii) $$

To calculate
$$\Delta H_a = ? $$

For $$2Al + Cr_2O_3\rightarrow Al_2O_3 + 2Cr .... (iv)$$

Subtract $$\Delta H_2$$ from $$\Delta H_1$$ to obtain $$\Delta H_a$$
$$\Delta H_a = \Delta H_1 - \Delta H_2$$
$$\Delta H_a= -1676 - (-1141)$$
$$\Delta H_a= -535 kJ$$

$$\Delta H_b = ? $$

For $$2Al + Fe_2O_3\rightarrow Al_2O_3 + 2Fe ......(v)$$

Subtract $$\Delta H_3$$ from $$\Delta H_1$$ to obtain $$\Delta H_b$$
$$\Delta H_b = \Delta H_1 - \Delta H_3$$
$$\Delta H_b= -1676 - (-822.2)$$
$$\Delta H_b= -853.8 kJ$$

For reaction (iv)

$$2\times 27g$$ of $$Al$$ produces 535 kJ of heat

10g of $$Al$$ produces $$ \dfrac { 535\times 10 }{ 2\times 27 } = 99.07 kJ$$

For reaction (v)

$$2\times 27g$$ of Al produces 853.8 kJ of heat

10g of Al produces $$\dfrac{ 853.8\times 10 }{ 54 }$$
$$ = 158.11 kJ$$

Reaction (ii) produces more heat by $$ = 158.11 - 99.07$$
$$ = 59.04 kJ$$
Question 8
1 / -0

If $$2Al(s) + 1\frac { 1 }{ 2 }O_2(g)\longrightarrow Al_2O_3(s), \Delta H = -166.78 kJ$$
$$H_2(g) + \frac { 1 }{ 2 }O_2(g)\longrightarrow H_2O(l), \Delta H = -285.9 kJ$$

Then $$\Delta H$$ for the reaction,

$$Al_2O_3(s)\longrightarrow 2Al(s) + 1\frac { 1 }{ 2 }O_2(g)$$ is:

A
$$\Delta H = 174.20kJ mol^{ -1 }$$

B
$$\Delta H = 124.55 kJ mol^{ -1 }$$

C
$$\Delta H = 166.78 kJ mol^{ -1 }$$

D
$$\Delta H = 156.33 kJ mol^{ -1 }$$

Solution

As given,
$$2Al(s) + 1\frac { 1 }{ 2 }O_2(g)\longrightarrow Al_2O_3(s), \Delta H = -166.78 kJ$$
and by this we can see that,
for reverse reaction
$$Al_2O_3(s)\longrightarrow 2Al(s) + 1\frac { 1 }{ 2 }O_2(g) \Delta H = 166.78 kJ$$
Question 9
1 / -0

The heat of combustion of acetylene is 312 kcal. If heat of formation of $$CO_2$$ and $$H_2O$$ are 94.38 and 68.38 kcal respectively, $$C \equiv\!\! \equiv C$$ bond energy is $$x \:kcal $$. Given that heat of atomisation of C and H are 150.0 and 51.5 kcal respectively and $$C-\! \! \! -H$$ bond energy is 93.64 kcal. Value of $$x$$

A
161

B
150

C
170

D
None of these

Solution

Given,
$$C_2H_2 + (5/2)O_2\longrightarrow 2CO_2 + H_2O; \: \Delta H = -312 \:kcal$$   ... (i)
                                       ( $$\because $$ Heat of combustion are exothermic)

$$C+ O_2\longrightarrow CO_2; \:    \Delta H = -94.38 \:kcal$$    ... (ii)

$$H_2+\frac{1}{2}O_2\longrightarrow H_2O;\: \Delta H = -68.38 \:kcal$$    ... (iii)

Multiply eq. (ii) by 2 and add in eq. (iii)

$$2C+ H_2+ (5/2)O_2\longrightarrow 2CO_2+ H_2O \: ; \Delta H = -257.14 \:kcal$$ ...( iv)

Subtracting eq. (i) from eq. (iv)

$$C_2H_2 + (5/2)O_2\longrightarrow 2CO_2 + H_2O    \: ; \Delta H = -312 \:kcal$$ ...( i)
-            -            -           -                                               +
........................................................................................................................
            $$2C+ H_2\longrightarrow C_2H_2; \:      \Delta H = +54.86 \:kcal$$         ... (v)

Also $$\Delta H$$ for $$C+ H_2\longrightarrow C_2H_2$$

$$\Delta H=Bond \: energy \: data \: of \: formation \: of \: bond + Bond \: energy \: data \: of \: dissociation \: of \: bond$$

$$ \Delta H= - [e _{(C \equiv\!\! \equiv C)} + 2\times e_{(C-\!\! \! -H)}]+[2C_{s\rightarrow g}+e_{(H-\!\! \! -H)}]$$

Also given, $$C_{s\rightarrow g}=150 \:kcal$$

$$\displaystyle \frac{1}{2}H_2(g)\longrightarrow H (g)- 51.5 \:kcal$$

$$\therefore          e_{H-\!\! \! -H}=51.5\times 2 \:kcal= 103.0 \:kcal$$

$$e_{C-\!\! \! -H}= 93.64 \:kcal$$

$$\therefore   54.86 = -[e _{(C \equiv\!\! \equiv C)}+2\times 93.64] +[2\times 150+1\times 103.0]$$

$$\therefore         e _{C \equiv\!\! \equiv C}=160.86 \:kcal$$
Question 10
1 / -0

A change in the free energy of a system at constant temperature and pressure will be:

$$\Delta_{ sys }G = \Delta_{ sys }H - T\Delta_{ sys }S$$

At constant temperature and pressure,
$$\Delta_{ sys }G < 0 (spontaneous)$$
$$\Delta_{ sys }G = 0 (equilibrium)$$
$$\Delta_{ sys }G > 0 (non-spontaneous)$$

For a system in equilibrium, $$\Delta G = 0$$, under conditions of constant_____

A
temperature and pressure

B
pressure and volume

C
temperature and volume

D
energy and volume

Solution

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Thermodynamics Test - 51

Result

Thermodynamics Test - 51

Score

-

Rank

-

SHARING IS CARING

Question 1

$$X= -108 $$

$$X= +108 $$

$$X= -59$$

None of these

Solution

Question 2

$$\Delta H= -78\ kcal$$

$$\Delta H= +78\ kcal$$

$$\Delta H= -498\ kcal$$

$$\Delta H= +498 \ kcal$$

Solution

Question 3

$$\ 1667.8\ kJ mol^{ -1 }$$

$$\ 1222.3\ kJ mol^{ -1 }$$

$$\ 1458.2\ kJ mol^{ -1 }$$

$$\ 1785.3\ kJ mol^{ -1 }$$

Solution

Question 4

$$\Delta H= - 850.2 kJ mol^{ -1 }$$

$$\Delta H= +850.2 kJ mol^{ -1 }$$

$$\Delta H= - 2500 kJ mol^{ -1 }$$

None of these

Solution

Question 5

$$\Delta H = -185.92 Kcal$$

$$\Delta H = +185.92 Kcal$$

$$\Delta H = 0 Kcal$$

None of these

Solution

Question 6

$$13.4$$

$$26.8$$

$$30.2$$

none of these

Solution

Question 7

Reaction (b) produces more heat by $$59.04 kJ$$

Reaction (a) produces more heat by $$59.04 kJ$$

Reaction(a) and (b) produces equal heat

None of these

Solution

Question 8

$$\Delta H = 174.20kJ mol^{ -1 }$$

$$\Delta H = 124.55 kJ mol^{ -1 }$$

$$\Delta H = 166.78 kJ mol^{ -1 }$$

$$\Delta H = 156.33 kJ mol^{ -1 }$$

Solution

Question 9

161

150

170

None of these

Solution

Question 10

temperature and pressure

pressure and volume

temperature and volume

energy and volume

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.