Use the data in Appendix B to calculate ∆G for the decom-position of nitrosyl chloride at 25 °C when the partial pressures are 2.00 atm of NOCl, 1.00 * 10^-3 atm of NO, and 1.00 * 10^-3 atm of Cl2: Is the reaction spontaneous in the forward or the reverse direction under these conditions?

Video transcript

Hello in this problem, we are asked to consider the gas equilibrium for the demonization of nitrogen dioxide at kelvin. Where asked what is the value for the change in the gibbs free energy under nonstandard conditions when nitrogen dioxide has a pressure 1.23 atmospheres and die, nitrogen tetroxide has a pressure of 2.45 atmospheres. Using the following change in gibbs free energy under standard conditions, values We want to know in which direction florida reverse is this reaction spontaneous under the given conditions. So recall that then our change in gibbs free energy under nonstandard conditions can be determined from the changing gibbs Free energy under standard conditions by the equation shown here where we have the change in Gibbs. Free energy under nonstandard conditions is equal to that understanding conditions plus the gas constant times temperature times the natural log of Q. Which is our reaction quotient. So our reaction quotient Q then is equal to the partial pressure of our product, the nitrogen tetroxide all over the partial pressure of our reactant, nitrogen dioxide squared. And so we're told the pressures of each of these in the problem statement. And so this works out then to 1.619, we can calculate the change in gibbs free energy under standard conditions is equal to the sum of the changing gibbs free energy of formation for the products minus the sum of the change in gibbs free energy for the reactant. So make use of those values that are provided. So we have then one more of our product times its standard gibbs free energy of formation. We subtract off. Then we have two moles of our reactant times its standard changing gibbs free energy. And so we have moles cancels and we're left with killed jules. This works out to the negative 2.8 kill jules. Well, then plug this information into our equation to find the change in gibbs strategy under nonstandard conditions. This is equal to then the change in strategy. Under standard conditions plus r gas constant times are temperature 350 Kelvin times the natural log of our reaction potion. So are changing gibbs free energy under Standard conditions works out to negative 1.4. Since this is negative in sign, it means then that the reaction will be spontaneous in the four direction. Thanks for watching. Hope this helps