Question

For the reaction at 298K, 2NO₂(g)$\rightleftharpoons {\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)$, the values of $∆\mathrm{H}°\mathrm{and}∆\mathrm{S}°\mathrm{are}-58.03\mathrm{kJ}\mathrm{and}-176.6\mathrm{J}/\mathrm{K},$, respectively. What is the value of $\mathbf{∆}\mathbf{G}\mathbf{°}$at 298 K? Assuming that $∆\mathrm{H}°\mathrm{and}∆\mathrm{S}°$do not depend on temperature, at what temperature is $∆\mathrm{G}°=0$? Is $∆\mathrm{G}°$ negative above or below this temperature?

Short answer

Answer:

Value of change in Gibbs free energy at 298K is -5.403kJ

The temperature at which change in Gibbs free energy is zero is equal to 328.6K

The value of $∆\mathrm{G}°$is negative below the temperature328.6K

Step-by-step solution

Definition of Gibbs free energy

The enthalpy of the system minus the product of the temperature times the entropy of the system is defined as Gibbs free energy

Calculation for the value of ∆G°  at  298 K and at what temperature is

$\begin{array}{rcl}∆\mathrm{G}°& =& 0\\ ∆\mathrm{H}& =& -58030\mathrm{J}\\ ∆\mathrm{S}°& =& -176.6\mathrm{J}/\mathrm{K}\\ ∆{\mathrm{G}}_{298}^{°}& =& -58030\mathrm{J}-\left(-298\mathrm{K}\times -176.6\mathrm{J}/\mathrm{K}\right)\\ ∆{\mathrm{G}}_{298}^{°}& =& -5.403\mathrm{kJ}\end{array}$

Value of change in Gibbs free energy at 298K is -5.403kJ

To calculate temperature, put the value of Gibbs free energy equals to zero

$\begin{array}{rcl}∆{\mathrm{G}}_{298}^{°}& =& ∆\mathrm{H}°-\mathrm{T}∆\mathrm{S}°\\ 0& =& ∆\mathrm{H}°-\mathrm{T}∆\mathrm{S}°\\ \mathrm{T}& =& \frac{∆\mathrm{H}°}{∆\mathrm{S}°}\\ \mathrm{T}& =& 328.6\mathrm{K}\end{array}$

The temperature at which change in Gibbs free energy is zero is equal to 328.6K

The value of$∆\mathrm{H}°$ and$∆\mathrm{S}°$ is negative. The value of$∆\mathrm{G}°$is also negative at low temperatures and process is spontaneous

The value of $∆\mathrm{G}°$ is negative below the temperature 328.6 K