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1

JEE Main 2021 (Online) 27th August Evening Shift

Numerical
Data given for the following reaction is as follows :

FeO(s) + C(graphite) $$\to$$ Fe(s) + CO(g)

Substance $$\Delta H^\circ $$
(kJ mol$$^{ - 1}$$)
$$\Delta S^\circ $$
(J mol$$^{ - 1}$$ K$$^{ - 1}$$)
$$Fe{O_{(s)}}$$ $$ - 266.3$$ 57.49
$${C_{(graphite)}}$$ 0 5.74
$$F{e_{(s)}}$$ 0 27.28
$$C{O_{(g)}}$$ $$ - 110.5$$ 197.6


The minimum temperature in K at which the reaction becomes spontaneous is ___________. (Integer answer)
Your Input ________

Answer

Correct Answer is 964

Explanation

$${T_{\min }} = \left( {{{{\Delta ^0}H} \over {{\Delta ^0}S}}} \right)$$

$${\Delta ^0}{H_{rxn}} = \left[ {\Delta _f^0H(Fe) + \Delta _f^0H(CO)} \right] - $$

$$ = \left[ {\Delta _f^0H(FeO) + \Delta _f^0H({C_{(graphite)}})} \right]$$

$$ = [0 - 110.5] - [ - 266.3 + 0]$$

= 155.8 kJ/mol

$${\Delta ^0}{S_{rxn}} = \left[ {{\Delta ^0}S(Fe) + {\Delta ^0}S(CO)} \right] - $$

$$\left[ {{\Delta ^0}S(FeO) + {\Delta ^0}S({C_{(graphite)}})} \right]$$

$$ = [27.28 + 197.6] - [57.49 + 5.74]$$

= 161.65 J/mol-K

$${T_{\min }} = {{155.8 \times {{10}^3}J/mol} \over {161.65J/mol - K}} = 963.8$$ K

$$ \simeq 964$$ k (nearest integer)
2

JEE Main 2021 (Online) 27th August Evening Shift

Numerical
Two flasks I and II shown below are connected by a valve of negligible volume.


When the valve is opened, the final pressure of the system in bar is x $$\times$$ 10$$-$$2. The value of x is __________. (Integer answer)

[Assume - Ideal gas; 1 bar = 105 Pa; Molar mass of N2 = 28.0 g mol$$-$$1; R = 8.31 J mol$$-$$1 K$$-$$1]
Your Input ________

Answer

Correct Answer is 84

Explanation

Applying; (nI + nII)initial = (nI + nII)final

$$\Rightarrow$$ Assuming the system attains a final temperature of T (such that 300 < T < 60)

$$\Rightarrow$$ $$\left( {\matrix{ {Heat\,lost\,by} \cr {{N_2}\,of\,container} \cr I \cr } } \right) = \left( {\matrix{ {Heat\,gained\,by} \cr {{N_2}\,of\,container} \cr {II} \cr } } \right)$$

$$\Rightarrow$$ n1Cm(300 $$-$$ T) = nIICm(T $$-$$ 60)

$$ \Rightarrow \left( {{{2.8} \over {28}}} \right)(300 - T) = {{0.2} \over {28}}(T - 60)$$

$$\Rightarrow$$ 14(300 $$-$$ T) = T $$-$$ 60

$$\Rightarrow$$ $${{(14 \times 300 + 60)} \over {15}} = T$$

$$\Rightarrow$$ T = 284 K (final temperature)

$$\Rightarrow$$ If the final pressure = P

$$\Rightarrow$$ (nI + nII)final = $$\left( {{{3.0} \over {28}}} \right)$$

$$\Rightarrow$$$${P \over {RT}}({V_I} + {V_{II}}) = {{3.0\,gm} \over {28\,gm/mol}}$$

$$P = \left( {{3 \over {28}}mol} \right) \times 8.31{J \over {mol - K}} \times {{284K} \over {3 \times {{10}^{ - 3}}{m^3}}} \times {10^{ - 5}}{{bar} \over {Pa}}$$

$$\Rightarrow$$ 0.84287 bar

$$\Rightarrow$$ 84.28 $$\times$$ 10$$-$$2 bar

$$\Rightarrow$$ 84
3

JEE Main 2021 (Online) 27th August Morning Shift

Numerical
200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is $$-$$57.1 kJ. The increase in temperature in $$^\circ$$C of the system on mixing is x $$\times$$ 10$$-$$2. The value of x is ___________. (Nearest integer)

[Given : Specific heat of water = 4.18 J g$$-$$1 K$$-$$1, Density of water = 1.00 g cm$$-$$3]

[Assume no volume change on mixing)
Your Input ________

Answer

Correct Answer is 82

Explanation

$$\Rightarrow$$ Millimoles of HCl = 200 $$\times$$ 0.2 = 40

$$\Rightarrow$$ Millimoles of NaOH = 300 $$\times$$ 0.1 = 30

$$\Rightarrow$$ Heat released = $$\left( {{{30} \over {1000}} \times 57.1 \times 1000} \right)$$ = 1713 J

$$\Rightarrow$$ Mass of solution = 500 ml $$\times$$ 1 gm/ml = 500 gm

$$\Rightarrow$$ $$\Delta T = {q \over {m \times c}} = {{1713J} \over {500g \times 4.18{J \over {g - K}}}}$$ = 0.8196 K

= 81.96 $$\times$$ 10$$-$$2 K
4

JEE Main 2021 (Online) 26th August Evening Shift

Numerical
For water $$\Delta$$vap H = 41 kJ mol$$-$$1 at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___________ kJ mol$$-$$1

[Use : R = 8.3 J mol$$-$$1 K$$-$$1]
Your Input ________

Answer

Correct Answer is 38

Explanation

H2O(l) $$\to$$ H2O(g) : $$\Delta$$H = 41 $${{kJ} \over {mol}}$$

$$\Rightarrow$$ From the relation : $$\Delta$$H = $$\Delta$$U + $$\Delta$$ngRT

$$\Rightarrow$$ 41$${{kJ} \over {mol}}$$ = $$\Delta$$U + (1) $$\times$$ $${{8.3} \over {1000}}$$ $$\times$$ 373

$$\Delta$$ DU = 41 $$-$$ 3.0959 = 38 kJ/mol

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