Which of the following reactions has ∆Ssys> 0? (a) N2(g) + 3H2(g) -> 2NH3(g) (b) Ag+(aq) + Cl-(aq) -> AgCl(s) (c) 2H2O2(aq) -> 2H2O(l) + O2(g) (d) 2I(g) -> I2(g)

Video transcript

everyone's in this video. We want to go ahead and look at these reactions here, A 30. And to identify which one has an entropy. So delta S. Of the system that's greater than zero. Let's go ahead and recall what entropy even is. So entropy is the degree of chaos or disorder of a system. It depends on the following. So one of them is going to be the phase. So the gas phase has the highest entropy followed by our liquid and then the solid phase has the lowest entropy. Next thing is dependent on is going to be our complexity. So the more elements in the compound leads to a higher entropy. So as an example, we have N. 03 which has four atoms or four elements And this has a higher entropy than our, let's say N 02 gas to both gasses, this one has only three atoms or three elements. Next is dependent on the mass. So the greater the mass, the greater the entropy. So let's take into consideration our Xenon gas which has a greater mass and higher entropy. Then let's say our helium gas which has a lower mass. Alright, so let's take those into consideration when we're solving for this problem again, we identify which one has it all to s that's greater than zero. So basically a positive number. So for my first one here we have two moles of K C L. O three which is a solid going into two moles of K C L. Solid and three moles of 02 gas. So we have two solid reactant forming two solid products as well as three gashes products. So we're going to go from a solitary gas in this reaction, which means that entropy will increase because the randomness is increasing. So this means that my delta S. Is going to be a positive value. And moving on to statements or equation B. We're going from two moles of H two gas reacting with one mole of co two gas that leads to two moles of H 20 in its liquid state. So we can see here that we have three total cashes reactant that form to liquid products that were going from a gas to liquid. So gas to liquids, we're going backwards here and this means that our entropy will decrease because you know that gas has the highest entropy. So my delta S in this case is going to be a negative. Now moving on to statements or reactions seem we have two moles of magnesium solid reacting with one mole 02 gas that gives us two moles of our M. G. O. Solid. So we have two solid reactions and one gas reactant that forms to solid products. So we're going from a gas to a solid, you know, solid has the least entropy and gas hoses most. So we're going backwards and because this means that our entropy will decrease, which means that my delta S is going to be equal to well it's going to be having a sign of negative for my last statement here we have three moles of C. Two plus and two moles of p 43 minus two. Give us a solid product. So we have to Aquarius reactions forming one solid product so that this phase is technically a liquid. So we're gonna go ahead and consider this going From a liquid to a solid race that the solid has the least amount of entropy. So we're going to go ahead and decrease in our value, meaning that we will have a negative value. So my dealt to us is of course negative. So the only one from a 30 that has a entry piece without us, that square than zero is going to be statement a. So A is going to be my final answer for this problem. Thank you so much for watching.