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
MCQ (Single Correct Answer)
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
MCQ (Single Correct Answer)
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
MCQ (Single Correct Answer)
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 .
Questions Asked from Thermodynamics
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