The lead storage battery uses the reaction: (b) Calculate ∆G for this reaction on a cold winter's day (10 °F) in a battery that has run down to the point where the sulfuric acid concentration is only 0.100 M.

Video transcript

Hello everyone today. We have the following problem. The chemical reaction that takes place at the and node of a nickel cadmium battery is as follows, calculate the value of our change in the gibbs free energy for this reaction at negative 15 degrees Celsius, assuming that the battery is discharged to a point where the hydroxide ion concentration has dropped 2.0 to molar. So first we need to establish our relationship between our gifts for energy or changing its for energy. And this gives free energy value here of the system. So we have the gibbs free energy or the change is equal to the changeup gives free energy within the system plus our gas constant times our temperature times the natural log of our Q constant. So essentially we need to find the value of Q and Q. Can be found by the concentration of our products divided by the concentration of our reactant. For this reaction. Solids are not considered within this equilibrium expression. So we're just gonna simply use a one for reactant. We just have our hydroxide or two moles. So that's going to be our hydroxide and print and brackets with a two as an exponent. And that comes from the X. The coefficient that's in front of the hydroxide. So we plug in our unit our values, we have one over our concentration of hydroxide units which is 0.025 moller we square that we get a q value of 1600. Before we plug in our final values, we need to convert our temperature from C to Kelvin. So we have negative 15°C and we're simply going to add 273.15 to get to 58.15 Kelvin. So now we can plug in our values. So we had our change in our gibbs. Free energy of the system being negative 1 59. kg per mole plus our gas constant. Which I'm actually going to move this down here for more space we have a negative 59.2 kilograms per mole Plus our gas constant which is 8.3 times 10 to the negative third. Killing joules per mole, kelvin Times are temperature which is 285.15 Kelvin Times are natural log of our Q which was 1600. This is going to result in a final gibbs. Free energy change of negative 143.4 kg per mole as our final answer. And without the of answer the question overall, I hope this helped. And until next time