1
JEE Main 2018 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
The combustion of benzene(l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 25oC; heat of combustion (in kJ mol–1) of benzene at constant pressure will be :
(R = 8.314 JK–1 mol–1)
A
–3267.6
B
4152.6
C
–452.46
D
3260
2
JEE Main 2018 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction? JEE Main 2018 (Offline) Chemistry - Thermodynamics Question 149 English
A
A and D
B
A and B
C
B and C
D
C and D
3
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
4
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
$$\Delta $$fGo at 500 K for substance 'S' in liquid state and gaseous state are +100.7 kcl mol-1 and +103 kcal mol-1, respectively. Vapour pressure of liquid 'S' at 500 K is approximately equal to : ( R = 2 cal K-1 mol-1 )
A
0.1 atm
B
1 atm
C
10 atm
D
100 atm
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