Question

For the following reaction : 2 N205 (g) → 4 NO2 (g) + O2 (g),
the standard Gibbs free energy change is -29.0 kJ/mol. What is the value of Gibbs free energy
change for this reaction at 350 K when [N2051 = 0.602 M, [NO2] = 0.305 M, and [O2] = 2.10 x 10-2 M
?​

Answer 1

Answer:

what is this in 2050 for 2032 Hafiz saide the

Answer 2

Given:

for reaction, 2 N205 (g) → 4 NO2 (g) + O2 (g)

the standard Gibbs free energy change( ΔG$_{0}$) = -29.0 KJ/mol

temperature = 350K

[N2O5] = 0.602M

[NO2] = 0.305 M

[O2] = 2.10 × 10-2 M

To Find:

the value of Gibbs free energy (ΔG ) =?

Solution:

We know that, for calculating K$_{eq}$ of reaction,

$K_{eq} = \dfrac{[NO_{2}]^{4} \times [O_{2} ]}{[N_{2}O_{5}]^{2} }$

we know the value of [N2O5] =0.602M, [NO2] =0.305 M, [O2] =2.10 × 10-2M

On putting values in above equation,

K$_{eq}$ = 0.50 × 10-2

now, for change in gibbs energy,

$\delta G = \delta G_{o} + RT lnK_{eq}$

ΔG = -29000 + 8.314 × 350 × ln 0.50 × 10-2

     on solving above equation,

ΔG = 44416.65 J/mol = 44.416 kJ/mol

Hence, the value of Gibbs free energy is 44.42 kJ/mol