Question

3
At what temperature is the following reaction at equilibrium? 2Na_2O_2+2H_2O right arrow 4NaOH+O_2 delta Hdegree=-109 kJ, deltaSdegree=-133.2j/K

Answer 1

Here we have the reaction

$2Na_2O_2+2H_2O\rightarrow 4NaOH+O_2$

And

$\\ \Delta H^\circ =-109KJ=-109000J\\ \Delta S^\circ = -133.2J/K$

At equilibrium free energy is zero that means $\\ \Delta G^\circ =0$ ,

As we know

$\\ \Delta G^\circ =\Delta H^\circ -T\Delta S^\circ \\ \text{At equilibrium, we have }\\ \Delta G^\circ =0\\ \Delta H^\circ -T\Delta S^\circ=0\\ \Delta H^\circ=T\Delta S^\circ\\ T=\frac{\Delta H^\circ}{\Delta S^\circ}\\ T= \frac{-109KJ}{-133.2J/K}\\ T= \frac{-109\times 1000J}{-133.2J/K}\\ T= 818.31 K$

Hence the temperature required at equilibrium is 818.31 K.