Consider the unbalanced equation: (b) Use the data in Appendix B and ΔG°f for IO₃^-(aq)= -128.0 kJ/mol to calculate ΔG° for the reaction at 25 °C.

Video transcript

Hello everyone today. We have the following problem. Consider the disproportion ation of mercury chloride under UV light. Use the following data to calculate the value the value of our Gibbs. Free energy for this reaction at one atmospheres and at 25°C. So let's recall the equation or the formula for our Gibbs. Free energy for reaction. It's going to be the sum of the gibbs. Free energy of our products minus the sum of our gifts. Free energy of our reactant. So what is that gonna look like when we write it out? Well we see That we have one mole of our mercury. And when we multiply it we're going to get zero Because it's gibbs free energy is zero. We are then going to add that to our one more of our mercury chloride. But this time we can multiply this by our negative 1 78. kg joules primal. And lastly we're going to subtract because those were our products from our reactant which associated which was our mercury chloride. And we had one mole of that. And then we multiplied it by its gibbs. Free energy change Which was negative 2 10. kg per mole. When we solve for our Gibbs free energy for this reaction, we're gonna get 32 .1 as our final answer. Overall I hope this helped until next time