Question

consider the following reaction 2no(g)+o2(g)→2no2(g). calculate ∆g° at 298k and predict weather the reaction is spontaneous or not ∆g° (no) is 86.69kj/mol ,∆g°(no2) is 51.84kj/mol​

Answer 1

Explanation:

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Answer 2

The value of $\Delta G^o$ is, -70.24 kJ and the given reaction is spontaneous.

Explanation :

The given balanced chemical reaction is,

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

Now we have to calculate the value of $(\Delta G^o)$.

$\Delta G^o=G_{product}-G_{reactant}$

$\Delta G^o=[n_{NO_2(g)}\times \Delta G_f^0_{(NO_2(g))}]-[n_{NO(g)}\times \Delta G^0_{(NO(g))}+n_{O_2(g)}\times \Delta G^0_{(O_2(g))}]$

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(NO(g))}=86.69kJ/mol\\\Delta G^0_{(NO_2(g))}=51.84kJ/mol\\\Delta G^0_{(O_2(g))}=0kJ/mol$

Now put all the given values in this expression, we get:

$\Delta G^o=[2mole\times (51.84kJ/mol)]-[2mole\times (86.96kJ/mol)+1mole\times (0kJ/mol)]$

$\Delta G^o=-70.24kJ$

As we know that:

$\Delta G$= +ve, reaction is non spontaneous

$\Delta G$= -ve, reaction is spontaneous

$\Delta G$= 0, reaction is in equilibrium

From this we conclude that the given reaction is spontaneous because the value of $\Delta G^o$ is negative.

Hence, the value of $\Delta G^o$ is, -70.24 kJ and the given reaction is spontaneous.

Learn more about : change in Gibbs free energy and spontaneity.

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