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JEE Chemistry Solved Question Paper 2007

Multiple Choice Questions

When a solid melts reversibly :

H decreases
G increases
E decreases
S increases

2.
Enthalpy is equal to :

$T^{2} {[\frac{\partial (G / T)}{\partial T}]}_{P}$

$- T^{2} {[\frac{\partial (G / T)}{\partial T}]}_{P}$

$T^{2} {[\frac{\partial (G / T)}{\partial T}]}_{V}$

$- T^{2} {[\frac{\partial (G / T)}{\partial T}]}_{V}$

$- T^{2} {[\frac{\partial (G / T)}{\partial T}]}_{P}$

$The Gibbs - Helmholtz equation is as : G = H + T {(\frac{\partial G}{\partial T})}_{P} Dividing above equation by T^{2} \frac{G}{T^{2}} = \frac{H}{T^{2}} + \frac{1}{T} (\frac{\partial G}{\partial T}) This on rearrangement becomes {[\frac{\partial (G / T)}{\partial T}]}_{P} = \frac{H}{T^{2}} H = - T^{2} {[\frac{\partial (G / T)}{\partial T}]}_{P} Where H = enthalpy$

Condition for spontaneity in an isothermal process is

$∆ A + W < 0$
$∆ G + U < 0$
$∆ A + U > 0$
$∆ G - U < 0$

$2 C (s) + 2 O_{2} (g) \to 2 {CO}_{2} (g); ∆ H = - 787 kJ . . . . (i) H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); ∆ H = - 286 kJ . . . (ii) C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) \to 2 {CO}_{2} (g) + H_{2} O (l); ∆ H = - 1310 kJ . . . (iii)$

The heat of formation of acetylene is :

-1802 kJ
+1802 kJ
+237 kJ
-800 kJ

Given the equilibrium system:

${NH}_{4} Cl (s) \to {NH}_{4}^{+} (aq) + {Cl}^{-} (aq) (∆ H = + 3.5 kcal / mol) .$

What change will shift the equilibrium to the right?

Decreasing the temperature
Increasing the temperature
Dissolving NaCl crystals in the equilibrium mixture
Dissolving NH₄NO₃ crystals in the equilibrium mixture

Equivalent amounts of H₂ and I₂, are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to HI, the K_c at this temperature is :