Use data from Appendix IIB to calculate the equilibrium constants at 25 °C for each reaction. a. 2 CO( g) + O2( g) ⇌ 2 CO2( g)

Video transcript

Hi everyone here we have a question asking us using the free energy, what is the equilibrium constant for the reaction at 25°C to sulfur dioxide gasses plus oxygen gasses forms to sulfur trioxide gasses. So first we have to calculate our change in reaction energy. So our change in energy equals the change in energy of the product minus the change in energy of the reactant. So now we need to plug in our numbers. So our change in energy for the reaction equals two moles of sulfur trioxide Times negative. 371.1 killed joules per mole minus two moles of sulfur dioxide Times -300 .1 Kill jules per mole plus one mole of oxygen. Time zero killed jules per mole And that equals negative 142 killed jules per mole. So now we have to calculate for R K. So our free energy of our reaction equals negative gas constant times temperature times natural log of K. So K. Equals E. To the negative change in energy divided by r. Gas constant times are temperature, which equals E. To the negative negative 140 to kill jules, times one times 10 to the third, jewels divided by one. Kill a jewel. And that's to get it into jewels Divided by 8. Times 298. Kelvin And that equals 7.56 times 10 to the 24th. And that is our final answer. Thank you for watching. Bye