Use data from Appendix IIB to calculate the equilibrium
constants...

Question

Use data from Appendix IIB to calculate the equilibrium
constants at 25 °C for each reaction.
b. 2 H2S( g) ∆ 2 H2( g) + S2( g)

Accepted Answer

Welcome back everyone in this example, we have the reaction of die phosphorus and sorry about that wrong pen. So die phosphorus gas is converted into phosphene gas from the reaction where we have di phosphorus gas reacting with three moles of hydrogen gas to produce two moles of phosphate gas at 298 Kelvin. And we need to determine what the equilibrium constant for this reaction is going to be. They give us our free energy formation of each of our re agents for the reaction given. So we're going to begin by finding this standard free energy for the reaction. And from that value, we're going to plug it into the falling equation where we should recall that our standard free energy is equal to negative one times r gas constant R times r temperature in kelvin times the natural log of K. Which we should recall is our equilibrium constant. Which is what we need to solve for for this example. So let's begin again by finding this term for the standard free energy change. So, using our standard free energy of formation, we're going to recall that. Our standard free energy change of our reaction is calculated from the standard free energy change of our products. Subtracted from the standard free energy change of our reactant. And so we're going to plug in the quantities from above to solve for our standard free energy change of the reaction. So we're going to say that our standard free energy change of our reaction is equal to the free energy change of our products. Beginning with our only product which is our two moles of phosphate gas Multiplied by phosphate gasses. Free energy change of formation being 13.4 kg joules per mole. So that completes the sum of our free energy change of formation of our products. And now we want to subtract this from the free energy change of formation of our reactant. Beginning with our first reactant, which is our di phosphorus. So we have one mole of di phosphorus multiplied by die phosphorus is free energy change of formation given from the prompt as 103.7 kill a jules per mole. We're then going to add this to the free energy change of our second reactant, which will just continue below here with which is our three moles of hydrogen gas, multiplied by hydrogen gas is free energy change of formation given in the prompt as zero killer jewels. Permal because that is hydrogen in its standard state as a diatonic molecule. And this completes to some of our free energy change of formation of our reactant. And now we want to simplify our brackets. So we would say that our free energy change of our reaction is equal to Our blue brackets. Should simplify to 26.8 kg jewels because we would cancel out the term moles in both of our brackets. And then our brackets in red are going to simplify too one oh 3.7 kg Joel's where when we take the difference here we get the free energy change. The standard free energy change of our reaction equal to negative 76.9 kg joules per mole. I'm sorry. We did cancel out our unit small. So we just have killed jules and actually to go back. It doesn't appear that we need to cancel out our units of moles because recall that free energy change should actually be with units of kilo jewels Permal. So we're going to make that correction here. We were actually originally right. And this is going to be our free energy change for our reaction, which will plug into our formula to solve for the equilibrium constant. So we want to reorder this formula here so that we can say that and I'll just rewrite it down here, we have our free energy change of our reaction equal to negative one times the gas constant. R times temperature, times the natural log of our equilibrium constant. And so we're going to first begin by dividing both sides by negative R. T. So that it cancels out on the right hand side. And we would be able to simplify to delta G degree divided by negative R. T. Equal to the natural log of our equilibrium constant. Which we can cancel out the natural log term by taking each side to eulogize number as an exponents. And so we would say that our equilibrium constant is equal to Mueller's number. Where we have in our exponents are free energy change of our reaction divided by negative one times r gas constant R times our temperature in kelvin and plugging in. What we know, we would say that our equilibrium constant is equal to E. To the In our numerator. We plug in our standard free energy change, which we just found to be negative 76.9 kg jewels Permal. And in our denominator we have our gas constant. R multiplied by negative one, which we should recall is 8.314 with units of joules divided by moles times Calvin. This is then multiplied by our temperature in Kelvin given in the prompt as 298 Kelvin. And because we recognize that our gas constant R. Has the unit jewels. We expanded our numerator so that we can fit in a conversion factor to go from killer jewels. Two jewels by recalling that are prefix kilo, tells us that we have 10 to the third power of our base unit jewels, allowing us to cancel out killer jewels as well as jewels with jewels in the denominator we can also get rid of moles with moles in the dominator. And we can also get rid of kelvin in the denominator and this leaves us with no units at all for our equilibrium constant, which is actually okay because this is what it should be like for a constant, we shouldn't have units for equilibrium constant. Sorry, this is an exponent to Ulis number. So in our calculators we're going to carefully type this quotient and we should yield a result of 3.188 times 10 to the 13th power. And again we have no units left. Where when we analyze our calculation for sig figs, we can see that our results from our free energy change only had one decimal place. And so therefore our final result Should have a minimum of one sig fig Or sorry, a maximum of one sick fig. And so we would round this answer to about three Times 10 to the 13th power as our equilibrium constant. So this is our final answer for the equilibrium constant. I hope that everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

Use data from Appendix IIB to calculate the equilibrium constants at 25 °C for each reaction. b. 2 H2S( g) ∆ 2 H2( g) + S2( g)

Watch next