1
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Given

(i)   2Fe2O3(s) $$ \to $$ 4Fe(s) + 3O2(g);

$$\Delta $$rGo = + 1487.0 kJ mol-1

(ii)   2CO(g) + O2(g) $$ \to $$ 2CO2(g);

$$\Delta $$rGo = $$-$$ 514.4 kJ mol-1

Free energy change, $$\Delta $$rGo for the reaction

2Fe2O3(s) + 6CO(g) $$ \to $$ 4Fe(s) + 6CO2(g) will be :
A
$$-$$ 112.4 kJ mol-1
B
$$-$$ 56.2 kJ mol-1
C
$$-$$ 168.2 kJ mol-1
D
$$-$$ 208.0 kJ mol-1
2
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
At a certain temperature in a $$5$$ $$L$$ vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction,
         CO + Cl2 $$\rightleftharpoons$$ COCl2
At equilibrium, if one mole of CO is present then equilibrium constant (Kc) for the reaction is :
A
2
B
2.5
C
3
D
4
3
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The correct order of electron affinity is :
A
F > Cl > O
B
F > O > Cl
C
Cl > F > O
D
O > F > Cl
4
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are Mx and My, respectively where Mx = $${3 \over 4}$$ My. The relative lowering of vapor pressure of the solution in X is ''m'' times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of the solvent, the value of ''m'' is :
A
$${4 \over 3}$$
B
$${3 \over 4}$$
C
$${1 \over 2}$$
D
$${1 \over 4}$$
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