Use your answer from Problem 7.54 to calculate the following:

[O...

Question

Use your answer from Problem 7.54 to calculate the following:

[O2] at equilibrium when [CO2] = 0.18 mol/L and [CO] = 0.0200 mol/L

Accepted Answer

Welcome back, everyone. We need to consider the given equilibrium reaction where hydrogen gas reacts with iodine gas in equilibrium with two moles of hydrogen iodine gas. Calculate our concentration of iodine at equilibrium. When our concentration of hydrogen is equal to 20.1 moles per leader. And our concentration of hydrogen iodine is equal to 0.3 moles per liter with an equilibrium constant K equal to 2.90 times 10 to the negative first power at 300. Kelvin. Let's recall that our equilibrium constant K describes our concentration of products race to their geometric coefficient as an exponents divided by our concentration of reactant raised to their psychometric coefficients as an exponents. We're given our equilibrium constant K as the value 2.90 times 10 to the negative first power. And it's equal to in our numerator, our concentration of our gaseous reactant or sorry are gaseous product being our hydrogen iodine gas where we have a concentration given of 0.03 20 moles per liter. And this is raised to its psychometric coefficient from our given bounce equation as to Then in our denominator, we have our concentration of our reactant. Our first reactant is going to be our hydrogen gas where we have a given concentration of 0.0160 moles per liter. And it has a psychometric coefficient of one. So we don't need to write that in. And this is multiplied by what we're solving for, which is our concentration of iodine gas, our second reactant. So we'll place our denominator in parentheses. So we're going to begin by multiplying both sides by our concentration of hydrogen. 0.0160 moles per liter times are concentration of iodine that's going to cancel out those terms in our denominator. On the right hand side and on the left hand side, we multiply by the same thing, our concentration of hydrogen 0.1 60 most per leader times our concentration of iodine. This is going to give us in our next line 2.90 times 10 to the negative first power, our equilibrium constant multiplied by our concentration of 0. 60 moles per liter, multiplied by our concentration of iodine equal to our right hand side, which is just zero. Sorry, our concentration of hydrogen gas being 0.3 to zero most per liter, which is squared. So isolating now for our concentration of iodine on the left, we're going to divide both sides by 2.90 times 10 to the negative first power Times 0.0160 moles per leader so that it will cancel out those terms on the left. We're going to divide by that, those same terms on the right 2.90 times 10 to the negative first power times 0.1 60 moles per liter. And in our final line, we'll find that our concentration of iodine is equal to In our numerator and sorry, just to make a correction, this should be zero 03 to zero in our numerator here or rather on the right hand side in our previous step. So 0.3 to zero squared is going to give us the results. Now being 0.10 to four in our numerator, this is divided by our denominator where we have 2.90 times 10 to the negative first power. And actually let's just show what that product is. So the product of 2.90 times 10 to the negative first power times 0. 60 moles per liter gives us the value 4.64 times 10 to the negative third power in our denominator. And notice that our units of moles per liter will cancel out when we divide these terms. So we no longer need our concentration symbol in our fractions. We just have parentheses around our values. And we'll find that our concentration of iding gas is equal to the value of about 0.221 moles per liter since this is a concentration. So this would be our final answer highlighted in yellow as our concentration of iodine gas corresponding to choice C in the multiple choice. So I hope that this made sense and let us know if you have any questions.

Use your answer from Problem 7.54 to calculate the following: [O2] at equilibrium when [CO2] = 0.18 mol/L and [CO] = 0.0200 mol/L

TextbookSolutions.keyConceptsTitle

Equilibrium Constant (K)

Le Chatelier's Principle

Stoichiometry

Watch next