So what we will do is we will raise E which is the inverse of natural log to the quantity of each side on this equation that we have here. But we need to be careful about here is we don't want the natural log of our concentration. When we do this calculation, we will get 4.5. So some simplification get the natural log of our initial concentration. And when we do this, we have to do a little bit of rearrangement. So I will get the natural log of my initial concentration, which we also found, and we will divide this by the concentration that we do not know yet that we're trying to find which is equivalent to RK times t. So now I have all the information that I need. That's the concentration that we're trying to find. So the equation that we will need or interest rate law is the natural log of my initial concentration divided by my concentration at a specific time, which is equivalent to my rate constant multiplied by time and important to note here that my A is actually Indian Quest. And the way we're going to find this concentration is using the integrated rate law for a first order Kandic reaction. So now was important is now we can actually have all of the information needed to find the concentration of Indian plots. And when you multiply this out, you will get 4000 500 seconds. So when we do this, our units of hours, within minutes we'll cancel. We know that in every one hour there is 60 minutes and in every one minute there is 60 seconds. So we start with 1.25 hours and this is again some dimensional analysis. So now what we need to dio this convert are given time of hours into seconds. So now we're also one step closer to finding our concentration. And when we do this, you get 0.316 polarity. And this is important because we were given middle leaders.
#Chemical equation balancer disproportion plus
And to find this I will simply to note concentration like this of indium Plus we will take the moles that we just found and we will divide it by our leaders of H CEO. And when we do this calculation you will get 0.158 moles of Indian chloride And now to find the concentration of Indian plus and remember, concentration is moles of solute divided by our leaders of solution. One more of Indian chloride is 150 0.271 grams per mole And this will give us moles because our grams well simply cancel. And to get those two moles, we simply do dimensional analysis. So we start with 22.38 grams of Indian chloride. So the next step that we must take is converting our mass of Indian chloride two moles. And upon doing this calculation, I get that my rate constant this 0.1 seconds in verse. And so when I do that, I get k equals 0.693 divided by half life which in our case, a 606 to 7 seconds. And so if I want to get the rate constant K by itself, all I have to do is rearrange this equation. Connects are half life, which is t 1/2 is equivalent to 0.693 divided by K, where K is our rate constant. And so, for any general equation, you involving first order.
And we're also given that the half life which is noted by T one has it's 606 and seven seconds. #S(VI+)+4xxO(-II)#.Okay, so we're given this reaction involving Indian quests and Indian solid, and our first test is to find the concentration of Indian plus after a given time. And we compare this with #SO_4^(2-)#, i.e. And so we got #S(VI+)# and #S(-II)#.the average oxidation state is still #S(+II)#. And this sulfur assumes THE SAME oxidation state as the oxygen it replaces, i.e. Just as a comment, I like to think of sulfurous acid as the SULFUR analogue of sulfuric acid, where ONE sulfur has replaced ONE oxygen to give #H_2S_2O_3#. What would you observe in this reaction? Well, probably (i) the precipitate of a fine white powder of elemental sulfur, and (ii) maybe the foul odour of #SO_2#. Looking at it again, we gots both LHS and RHS neutral, LHS and RHS, #2xxS#, #3xxO#, and #2xxH#.so balanced with respect to mass and charge. #underbrace(S_2O_3^(2-) +2H^(+))_("sulfurous acid"*H_2SO_3) rarr S +H_2O +SO_2#Īgain, if mass and charge ARE NOT BALANCED, then we CANNOT accept it as a representation of chemical reality. #2S_2O_3^(2-) +4H^(+) rarr 2S +2H_2O +2SO_2#Īnd of course we could halve this, as per Nam D.'s suggestion. We add #(i) + (ii)# to remove the electrons. Are they balanced? Don't trust my arithmetic. zerovalent sulfur, and oxidized to #stackrel(+IV)"SO"_2#. an average oxidation state of #(-II+VI)/2=+II#. Thiosulfate, #S_2O_3^(2-)# has formal #stackrel(-II)S# and #stackrel(+VI)S# oxidation states, i.e.