1
JEE Main 2017 (Offline)
+4
-1
$$\Delta$$U is equal to :
A
Isobaric work
B
C
Isothermal work
D
Isochoric work
2
JEE Main 2017 (Offline)
+4
-1
Given, $${C_{(graphite)}} + {O_2} \to C{O_2}(g)$$;

$${\Delta _r}{H^o}$$ = - 393.5 kJ mol-1

$${{\rm H}_2}(g)$$ + $${1 \over 2}{O_2}(g)$$$$\to {{\rm H}_2}{\rm O}(l)$$

$${\Delta _r}{H^o}$$ = - 285.8 kJ mol-1

$$C{O_2}(g)$$ + $$2{{\rm H}_2}{\rm O}(l) \to$$ $$C{H_4}(g)$$ + $$2{O_2}(g)$$

$${\Delta _r}{H^o}$$ = + 890.3 kJ mol-1

Based on the above thermochemical equations, the value of $${\Delta _r}{H^o}$$ at 298 K for the reaction

$${C_{(graphite)}}$$ + $$2{{\rm H}_2}(g) \to$$ $$C{H_4}(g)$$ will be :
A
+144.0 kJ mol–1
B
– 74.8 kJ mol–1
C
-144.0 kJ mol–1
D
+ 74.8 kJ mol–1
3
JEE Main 2016 (Online) 10th April Morning Slot
+4
-1
If 100 mole of H2O2 decompose at 1 bar and 300 K, the work done (kJ) by one mole of O2(g) as it expands against 1 bar pressure is :

2H2O2(l)    $$\rightleftharpoons$$    2H2O(l) + O2(g)

(R = 8.3 J K $$-$$1 mol$$-$$1)
A
62.25
B
124.50
C
249.00
D
498.00
4
JEE Main 2016 (Online) 9th April Morning Slot
+4
-1
For the reaction,
A(g) + B(g) $$\to$$ C(g) + D(g), $$\Delta$$Ho and $$\Delta$$So are, respectively, − 29.8 kJ mol−1 and −0.100 kJ K−1 mol−1 at 298 K. The equilibrium constant for the reaction at 298 K is :
A
1.0 $$\times$$ 10$$-$$10
B
1.0 $$\times$$ 1010
C
10
D
1
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