Thermodynamics Test

Result

Thermodynamics Test - 48

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

For the reaction : $$X_2O_4(l)\rightarrow 2XO_2(g)$$. $$\Delta U = 2.1\,kcal,\ \Delta S = 20\,cal\,K^{ -1 }\,at\ 300\,K$$.
Hence $$\Delta G$$ is:

A
$$-7\,kcal$$

B
$$-4.7\,kcal$$

C
$$+2.7\,kcal$$

D
$$-2.7\,kcal$$

Solution

As we know,

$$\Delta H = \Delta U + \Delta n_gRT$$

$$= 2.1+ 2\times 0.002\times 300 = 3.3\,kcal$$

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

$$= 3.3-300\times (0.02) = -2.7\,kcal$$
Question 2
1 / -0

$$C(s) + O_2(g) \rightarrow CO_2(g), \quad \Delta{H} = + 94.0 K cal.$$

$$CO(g) +\frac{1}{2} O_2(g) \rightarrow CO_2(g) ,\quad \Delta H = -67.7 K cal.$$

From the above reactions find how much heat (Kcal mole$$^{-1}$$) would be produced in the following reaction:

$$C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g)$$

A
20.6

B
26.3

C
44.2

D
161.7

Solution

The thermochemical reactions are as given below.

$$C(s) + O_2(g) \rightarrow CO_2(g) \Delta H = + 94.0 K cal.$$ ......(1)

$$CO(g) +\frac{1}{2} O_2(g) \rightarrow CO_2(g), \Delta H = -67.7 K cal$$ ......(2)

The reaction (2) is reversed and added to the reaction (1) to obtain the reaction $$C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g)$$.

The heat produced in this reaction is $$94.0-(-67.7)=161.7$$ Kcal/mole
Question 3
1 / -0

200 ml of 1M HCl is mixed with 400 ml of 0.5M NaOH. The temperature rise in the calorimeter was found to be 4.0oC. Water equivalent of calorimeter is 25g and the specific heat of the solution is 1 cal/mL/degree. If the theoritical heat of neutralization of a strong acid and strong base is −13.5 kcal, then the percentage error in this experiment while calculating the heat of neutralization is

A
7.4

B
3

C
5.6

D
4.5

Solution

Total volume $$200+400=600 ml$$
200 ml of 1 M acid $$=$$ 400 ml of 0.5 M $$NaOH$$
200 m. eq. of acid reacts with 200 m.eq. of base $$=\Delta H$$
Heat of neutralization $$=5\times \Delta H$$
Heat produced during neutralization $$=$$ heat taken up by calorimeter + solution
$$=m_{1}s_{1}\Delta T+m_{2}s_{2}\Delta T$$
$$=(25\times 1 \times 4)+(600\times 1 \times 4)=2500 cal$$
$$\therefore$$ Heat of neutralization $$=-5\times 2500$$$$=-12500 cal=-12.5 kcal$$
% error $$=(\dfrac{13.5-12.5}{13.5})\times 100=7.4$$
Question 4
1 / -0

0.16 g of methane was subjected to combustion at $$27^{\circ}C$$ in bomb calorimeter. The temperature of calorimeter system (including water) was found to rise by $$0.5^{\circ}C$$. If the heat of combustion of methane at constant volume and constant pressure is $$x$$ kJ/mole, find the value of $$x$$. The thermal capacity of the calorimeter system is $$17.7\ kJ\ K^{-1},\ R=8.314\ JK^{-1}\ mol^{-1}$$.

A
$$-890$$

B
$$890$$

C
$$450$$

D
$$-450$$

Solution

Heat given by burning of 0.16 g methane = heat taken up by calorimeter = $$17.7 \times 0.5 =8.85 \:kJ$$
$$\displaystyle \therefore$$ Heat given by burning of 16 g methane

$$=\dfrac{8.85\times 16}{0.16}=885 \:kJ$$

The burning is taking place at constant volume in a bomb calorimeter.
$$\therefore \Delta U= -885 \: kJ \: mol^{-1}$$
$$\because \Delta H= \Delta U+\Delta nRT$$

$$CH_4+ 2O_2\longrightarrow CO_2 + 2H_2O(l)$$;
$$\therefore \Delta n = -2$$
$$\therefore \Delta H= -885+(-2)\times 8.314\times 10 ^{-3} \times 300$$
$$= -889.98\: kJ \: mol^{-1}$$

Hence, the correct option is $$\text{A}$$
Question 5
1 / -0

For the hypothetical reaction:
$${ A }_{ 2 }(g)+{ B }_{ 2 }(g)\rightleftharpoons 2AB(g)$$

$${ \Delta }_{ r }{ G }^{ o }$$ and $${ \Delta }_{ r }{ S }^{ o }$$ is $$20 kJ/mol$$ and $$-20{ JK }^{ -1 }{ mol }^{ -1 }$$ respectively at $$200K$$. If $${ \Delta }_{ r }{ C }_{ p }$$ is $$20{ JK }^{ -1 }{ mol }^{ -1 }$$ then $${ \Delta }_{ r }{ H }^{ o }$$ at $$400K$$ is :

A
$$20\ kJ/mol$$

B
$$7.98\ kJ/mol$$

C
$$28\ kJ/mol$$

D
none of these

Solution

As we know,

$$\Delta G =\Delta H- T \Delta S; \Rightarrow 20=\Delta H -200\times \dfrac{-20}{1000}; \Rightarrow \Delta H = 16kJ/mol$$

Also,
$$\Delta H= \Delta_r C_p (T_2 - T_1); \Delta H = 16+20\times \dfrac{200}{1000} = 20kJ/mol$$
Question 6
1 / -0

Specific heat of substance at m.pt. (or b.pt. or for isothermal process) and specific heat of a substance during adiabatic change respectively are

A
0, 0

B
0, $$\infty$$

C
$$\infty$$, 0

D
$$\infty, \infty$$

Solution

The amount of heat required to raise the temperature of unit mass of a substance by one degree celsius is known as specific heat. It is given by the ratio $$\displaystyle \left ( \frac{\partial H}{\partial T} \right )$$
When at constant pressure, one phase of a substance is in equilibrium with other phase (during melting or boiling), the heat $$\partial H$$ is supplied to the system, and the temperature change $$\partial T$$ is zero as the heat supplied is utilized for the phase transition.
Hence, the specific heat is infinite. This is also true for the isothermal process.
During adiabatic process, no heat is supplied to system. The term $$\partial H$$ is zero. Hence the specific heat is zero.
Question 7
1 / -0

The heat of reaction for $$A + \frac{1}{2} O_2 \rightarrow AO$$ is $$-50$$ kcal/mol and $$AO + \frac{1}{2} O_2 \rightarrow AO_2$$ is $$100$$ kcal/mol.
The heat of reaction (in kcal/mol) for $$A + O_2 \rightarrow AO_2$$ will be:

A
$$-50$$

B
$$+50$$

C
$$+100$$

D
$$+150$$

Solution

The thermochemical reactions are given below:

$$A + \dfrac{1}{2} O_2 \rightarrow AO \ ; \Delta H =-50$$ kcal/mol

$$AO + \dfrac{1}{2} O_2 \rightarrow AO_2\ ; \Delta H =100$$ kcal/mol

The above two reactions are added to obtain the reaction as follows:

$$A + O_2 \rightarrow AO_2$$

Therefore, the heat of the reaction is $$-50+100=50$$ kcal/mol.
Question 8
1 / -0

Combustion of surose is used by aerobic organisms for providing energy for the life sustaining processes. If all the capturing of energy from the reaction is done through electrical process (non P-V work) then calculate maximum available energy which can be captured by combustion of $$34.2gm$$ of sucrose
Given : $$\Delta {{H}_{combustion}}(sucrose)=-6000\ kJ. {mol}^{-1}$$
$$\Delta {{S}_{combustion}}=180J/K.mol$$ and body temperature is $$300K$$

A
$$600kJ$$

B
$$594.6kJ$$

C
$$5.4kJ$$

D
$$605.4kJ$$

Solution

As we know,

$$\Delta G = \Delta H - T \delta S =-6000-300\times 0.18 = -6054\ kJ/mol$$

So, for 34.2gm (0.1 mol) of sucrose maximum available energy $$ \Delta G = 605.4\ kJ$$
Question 9
1 / -0

Find the forward$$\displaystyle \:\Delta _{r}U^{\circ}$$ at 300K for the reaction $$4HCl(g)+O_{2}(g)\rightarrow 2Cl_{2}(g)+2H_{2}O(g)$$ Assume all gases are ideal. Given $$\displaystyle \:H_{2}(g)+Cl_{2}(g)\rightarrow 2HCl(g)$$ $$\Delta _{r}H_{300}^{^{\circ}}= -184.5 kJ/mol$$
$$\displaystyle \:2H_{2}(g)+O_{2}(g)\rightarrow 2H_2O(g)$$ $$\Delta _{r}H_{300}^{^{\circ}}= -483 kJ/mol$$ (Use R= 8.3 J/mole)

A
111.5 kJ/mole

B
-109.01 kJ/mole

C
-111.5 kJ/mole

D
None

Solution

$$4HCl(g)+O_2(g) \longrightarrow 2Cl_2(g)+2H_2O(g)......(i)$$
$$H_2(g)+Cl_2(g) \longrightarrow 2HCl(g)....(ii)$$
$$2H_2(g)+O_2(g) \longrightarrow 2H_2O(g)....(iii)$$
As it can be clearly seen
$$Eq(i)=Eq(iii)-2Eq(ii)$$
Thus $$\Delta H_r^{o} (i)=\Delta H_r^{o}(iii)-2\Delta H_r^{o}(ii)$$
$$=-483-(2 \times -184)$$
$$=-114kJ/mol$$
$$\Delta U_r^{o}=\Delta H_r^{o}-\Delta n_gRT$$
Thus $$\Delta U_r^{o}=-114-(4-5) \times 0.0083 \times 300$$
$$ \Rightarrow \Delta U_r^{o}=-111.5kJ/mol$$
Question 10
1 / -0

In the Born-Haber cycle for the formation of solid common salt ($$Nacl$$), the largest contribution comes from

A
the low ionization potential of $$Na$$

B
the high electron affinity of $$Cl$$

C
the low $$\Delta H_{vap}$$ of $$Na(s)$$

D
the lattice energy

Solution

REF.Image

$$\Delta _{Hf^{\circ }}=\Delta H_{sub}+IE+\Delta H_{diss}+EA+U$$

$$\Delta _{Hf^{\circ }}=108+496+122-349-788=-411 kJ/mole$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Thermodynamics Test - 48

Result

Thermodynamics Test - 48

Score

-

Rank

-

SHARING IS CARING

Question 1

$$-7\,kcal$$

$$-4.7\,kcal$$

$$+2.7\,kcal$$

$$-2.7\,kcal$$

Solution

Question 2

20.6

26.3

44.2

161.7

Solution

Question 3

7.4

3

5.6

4.5

Solution

Question 4

$$-890$$

$$890$$

$$450$$

$$-450$$

Solution

Question 5

$$20\ kJ/mol$$

$$7.98\ kJ/mol$$

$$28\ kJ/mol$$

none of these

Solution

Question 6

0, 0

0, $$\infty$$

$$\infty$$, 0

$$\infty, \infty$$

Solution

Question 7

$$-50$$

$$+50$$

$$+100$$

$$+150$$

Solution

Question 8

$$600kJ$$

$$594.6kJ$$

$$5.4kJ$$

$$605.4kJ$$

Solution

Question 9

111.5 kJ/mole

-109.01 kJ/mole

-111.5 kJ/mole

None

Solution

Question 10

the low ionization potential of $$Na$$

the high electron affinity of $$Cl$$

the low $$\Delta H_{vap}$$ of $$Na(s)$$

the lattice energy

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.