1
MCQ (Single Correct Answer)

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
2
MCQ (Single Correct Answer)

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.
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 12th January Morning Slot

For diatomic ideal gas in a closed system, which of the following plots does not correctly describe the relation between various thermodynamic quantities ?
A
B
C
D

Explanation

CP does not changes with change in pressure.
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 12th January Evening Slot

The combination of plots which does not represent isothermal expansion of an ideal gas is –

A
A and D
B
B and D
C
B and C
D
A and C

Explanation

For isothermal process of ideal gas,

PV = constants = K

$$ \therefore $$  P = $${k \over v}$$

So, the graph between P and $${1 \over v}$$ is straight line passing through the origin. So, graph (A) is correct.

As PV = K so the P and V curve is hyperbola. So, graph (B) is wrong.

In isothermal PV = constant.

So, graph (C) is right.

We know internal energy (u) = $${f \over 2}$$ nRT. Internal energy is function of temperature (T) only. So, U does not change when volume(V) changes. So, graph (D) is wrong.

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