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1

### AIPMT 2006

The enthalpy and entropy change for the reaction:
Br2(l) + Cl2(g) $$\to$$ 2BrCl(g)
are 30 kJ mol$$-$$1 and 105 J K$$-$$1 mol$$-$$1 respectively.
The temperature at which the reaction will be in equilibrium is
A
300 K
B
285.7 K
C
273 K
D
450 K

## Explanation

$$\Delta$$G = $$\Delta$$H – T$$\Delta$$S

Now, at equilibrium $$\Delta$$G = 0

0 = $$\Delta$$H – T$$\Delta$$S

$$\Rightarrow$$ 0 = 30000 –T (105)

$$\Rightarrow$$ T = $${{30000} \over {105}}$$ = 285.7 K
2

### AIPMT 2006

Identify the correct statement for change of Gibb's energy for a system ($$\Delta$$Gsystem) at constant temperature and pressure.
A
If $$\Delta$$Gsystem < 0, the process is not spontaneous.
B
If $$\Delta$$Gsystem > 0, the process is spontaneous.
C
If $$\Delta$$Gsystem = 0, the system has attained equilibrium.
D
If $$\Delta$$Gsystem = 0, the system is till moving in a particular direction.

## Explanation

$$\Delta$$Gsystem < 0, process is spontaneous.

$$\Delta$$Gsystem = 0, process is in equilibrium.

$$\Delta$$Gsystem > 0, process is not spontaneous.
3

### AIPMT 2005

The absolute enthalpy of neutralisation of the reaction :

Mg(O)(s) + 2HCl(aq) $$\to$$ MgCl2(aq) + H2O(l) will be
A
$$-$$57.33 kJ mol$$-$$1
B
greater than $$-$$ 57.33 kJ mol$$-$$1
C
less than $$-$$ 57.33 kJ mol$$-$$1
D
57.33 kJ mol$$-$$1

## Explanation

We know that enthalpy of neutralization of a strong acid and a strong base is –57.33 kJ mol–1.

Here, MgO is the weak base and HCl is a strong acid thus, a small amount of energy is used in the ionization of MgO thus, the heat of neutralization decreases. Therefore, enthalpy of neutralization is less than – 57.33 kJ mol–1
4

### AIPMT 2005

A reaction occurs spontaneously if
A
T$$\Delta$$S < $$\Delta$$H and both $$\Delta$$H and $$\Delta$$S are +ve
B
T$$\Delta$$S > $$\Delta$$H and $$\Delta$$H is +ve and $$\Delta$$S are $$-$$ve
C
T$$\Delta$$S > $$\Delta$$H and both $$\Delta$$H and $$\Delta$$S are +ve
D
T$$\Delta$$S = $$\Delta$$H and both $$\Delta$$H and $$\Delta$$S are +ve

## Explanation

$$\Delta$$G = $$\Delta$$H – T$$\Delta$$S

For spontaneous reaction, $$\Delta$$G has to be negative.

Among the given options, it is positive only when T$$\Delta$$S > $$\Delta$$H and both $$\Delta$$H and $$\Delta$$S are +ve .

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