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1

### JEE Main 2019 (Online) 11th January Evening Slot

The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by $$\Delta$$rGo = A – BT

Where A and B are non-zero constants. Which of the following is TRUE about this reaction ?
A
Exothermic if B < 0
B
Endothermic if A > 0
C
Exothermic if A > 0 and B < 0
D
Endothermic if A < 0 and B > 0

## Explanation

$$\Delta$$Go = $$\Delta$$Ho - T$$\Delta$$So

Given that A and B are non-zero constants, i.e., A = $$\Delta$$Ho, B = $$\Delta$$So

If $$\Delta$$Ho is positive means reaction is endothermic.
2

### JEE Main 2019 (Online) 11th January Evening Slot

The reaction, MgO(s) + C(s) $$\to$$ Mg(s) + CO(g), for which $$\Delta$$rHo + 491.1 kJ mol–1 and $$\Delta$$rSo = 198.0 JK–1 mol–1, is not feasible at 298 K. Temperature above which reaciton will be feasible is :
A
2480.3 K
B
2040.5 K
C
2380.5 K
D
1890.0 K

## Explanation

We know,

$$\Delta$$Go = $$\Delta$$Ho - T$$\Delta$$So

For a reaction to be spontaneous $$\Delta$$Go must be negative i.e.,
T$$\Delta$$So > $$\Delta$$Ho

$$\Rightarrow$$ T > $${{\Delta H^\circ } \over {\Delta S^\circ }}$$

$$\Rightarrow$$ T > $${{491.1 \times 1000} \over {198}}$$

$$\Rightarrow$$ T > 2480.3 K
3

### JEE Main 2019 (Online) 11th January Morning Slot

Two blocks of the same metal having same mass and at temperature T1 and T2, respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, $$\Delta$$S, for this process is :
A
2Cp In $$\left[ {{{{{\left( {{T_1} + {T_2}} \right)}^{{1 \over 2}}}} \over {{T_1}{T_2}}}} \right]$$
B
2Cp In $$\left[ {{{\left( {{T_1} + {T_2}} \right)} \over {2{T_1}{T_2}}}} \right]$$
C
Cp In $$\left[ {{{{{\left( {{T_1} + {T_2}} \right)}^2}} \over {4{T_1}{T_2}}}} \right]$$
D
2Cp In $$\left[ {{{\left( {{T_1} + {T_2}} \right)} \over {4{T_1}{T_2}}}} \right]$$

## Explanation

When two blocks comes in contact with each other and attain thermal equilibrium then

final temperature of the blocks,

Tf = $${{{T_1} + {T_2}} \over 2}$$

$$\Delta$$S = $$\Delta$$S1 + $$\Delta$$S2

= Cp $$\ln \left( {{{{T_f}} \over {{T_1}}}} \right)$$ + Cp $$\ln \left( {{{{T_f}} \over {{T_2}}}} \right)$$

= Cp $$\ln \left( {{{{T_1} + {T_2}} \over {2{T_1}}}} \right)$$ + Cp $$\ln \left( {{{{T_1} + {T_2}} \over {2{T_2}}}} \right)$$

= Cp $$\ln \left( {{{{{\left( {{T_1} + {T_2}} \right)}^2}} \over {4{T_1}{T_2}}}} \right)$$
4

### JEE Main 2019 (Online) 11th January Morning Slot

For the chemical reaction X $$\rightleftharpoons$$ Y, the standard reaction Gibbs energy depends on temperature T (in K) as $$\Delta$$rGo (in kJ mol–1) = 120 $$- {3 \over 8}$$ T.

The major component of the reaction mixture at T is :
A
Y if T = 300 K
B
Y if T = 280 K
C
X if T = 350 K
D
X if T = 315 K

## Explanation

X $$\rightleftharpoons$$ Y

Keq = $${{\left[ Y \right]} \over {\left[ X \right]}}$$

If Keq > 1 $$\Rightarrow$$ [Y] > [X]

and Keq < 1 $$\Rightarrow$$ [Y] < [X]

We know, $$\Delta$$Go = -RT ln(Keq)

So when Keq > 1 then $$\Delta$$Go < 0 and Y is major.

And when Keq < 1 then $$\Delta$$Go > 0 and X is major.

Temperature at which $$Go = 0 is 120$$ - {3 \over 8}$$T = 0$$ \Rightarrow $$T = 320 K For T > 320 K$$\Delta $$Go < 0 and Y is major. For T < 320 K$$\Delta Go > 0 and X is major.

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