Question

Consider the following reaction at 298 K. C ( graphite ) + 2 H 2 ( g ) ⟶ CH 4 ( g ) Δ H ∘ = − 74.6 kJ and Δ S ∘ = − 80.8 J / K Calculate the following quantities. Δ S sys = J/K Δ S surr = J/K Δ S univ = J/K Is this reaction spontaneous? yes no

Answer 1

Answer :

Entropy of system = -80.8 J/K

Entropy of surrounding = 253.7 J/K

Entropy of universe = 172.9 J/K

The reaction is spontaneous.

The process is a spontaneous process.

Explanation :

The given chemical reaction is:

$C(graphite)+2H_2(g)\rightarrow CH_4(g)$

Entropy of reaction = $\Delta S^o$ = Entropy of system = -80.8 J/K

Now we have to calculate the entropy of surrounding.

Entropy of surrounding  = $\frac{-\Delta H^o}{T}$

Entropy of surrounding  = $-\frac{-75.6kJ}{298K}$

Entropy of surrounding  = $\frac{75.6\times 1000J}{298K}$

Entropy of surrounding  = 253.7 J/K

As, we know that:

Entropy of universe = Entropy of system + Entropy of surrounding

Entropy of universe = -80.8 J/K + (253.7 J/K)

Entropy of universe = 172.9 J/K

Now we have to calculate the Gibbs free energy.

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy  = ?

$\Delta H^o$ = standard enthalpy = -74600 J

$\Delta S^o$ = standard entropy = -80.8 J/K

T = temperature of reaction = 298 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(-74600J)-(298K\times -80.8J/K)$

$\Delta G^o=-50521.6J=-50.5kJ$

• A reaction to be spontaneous when $\Delta G<0$
• A reaction to be non-spontaneous when $\Delta G>0$

For the reaction to be spontaneous, the Gibbs free energy of the reaction $\Delta G$ is negative or we can say that the value of

As the value of $\Delta G$ is less than zero that means the reaction is spontaneous.