Question

For the reaction : [2014]
X₂O₄(????) ® 2XO₂(g)
∆U = 2.1 k cal, ∆S = 20 cal K⁻¹ at 300 K
Hence ∆G is:-
(a) 2.7 k cal (b) – 2.7 k cal
(c) 9.3 k cal (d) – 9.3 k cal

Answer 1

Explanation:

The relationship between the enthalpy change and the change in internal energy is as shown below.

\Delta H=\Delta U+\Delta n_gRTΔH=ΔU+Δn

g

RT

Substitute values in th above expression.

=2.1+\dfrac {2\times 2\times 300}{1000}=3.3=2.1+

1000

2×2×300

=3.3

The relationship between the change in free energy, the enthalpy change and the entropy change is as shown below.

\Delta G=\Delta H-T\Delta sΔG=ΔH−TΔs

=3.3-300\times \dfrac {26}{1000}=3.3-6=-2.7 K cal=3.3−300×

1000

26

=3.3−6=−2.7Kcal

Hence, the change in the free energy is -2.7 K cal.

Answer 2

The $\bold{\Delta\,G}$ of  the given reaction is -2.7 k cal.

Step by step explanation:

The given chemical reaction is as follows.

                             $\bold{X_{2}O_{4}\rightarrow 2XO_{2}}$

The mathematical expression for the enthalpy change and internal change is as follows.

                        $\bold{\Delta\,H=\Delta U+\Delta n_{g}RT}$.................(1)

Internal energy $\Delta U$ = 2.1 k cal

Temperature T = 300 K

From the reaction number of moles =2

Substitute the all given values in the equation (1)

                            $=2.1+\frac{2\times 2 \times 300}{1000}$

                            $\Rightarrow 3.3$

The relationship between the change in free energy, the enthalpy change and the entropy change is as follows.

                      $\bold{\Delta G = \Delta H - T \Delta S }$..........................(2)

Substitute the given values in equation (2)

                            $\Rightarrow 3.3-300\times \frac{26}{1000}$

                           $\Rightarrow 3.3-6$

                           $\Rightarrow -2.7K\,cal$

Therefore, change in free energy is -2.7 K cal.

Hence, correct option is -2.7 k cal.

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