Use your answer from Problem 7.54 to calculate the following:[O2]...

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. In this example, we need to consider the equilibrium reaction where hydrogen gass reacts with one mole of iodine gas and equilibrium to form two moles of hydrogen iodide gas. Calculate the concentration of iodine at equilibrium. When our concentration of hydrogen is equal to 0.0160 moles per liter. And the concentration of hydrogen iodide is equal to 0.0320 moles per liter where we're given our equilibrium constant K of 2.90 times 10 to the negative first power at 300. Kelvin. Let's begin by recalling that our equilibrium constant K is defined by our concentration of products raised to their stoch meric coefficient divided by the concentration of reactants raised to their stoch meric coefficient. So it's important that we observe whether our given reaction is balanced and based on the number of atoms on both sides of our reaction, we can confirm that our reaction as given in the prompt is balanced. So let's plug in what we know we're given our equilibrium constant K as 2.90 times 10 to the negative first power, which is equal to our concentration of our products where we have hydrogen iodide gas as 0.0320 moles per liter. This is then raised to its stom meric coefficient which is a two. Then we are dividing by our concentration of reactants where our first reactant is hydrogen gas, which has a concentration given as 0.0160 moles per liter, which is multiplied by our concentration of our second reactant iodine, which is what we need to solve for. And note that the to geometric coefficients are one as implied by our given balanced reaction. Now we're going to need to isolate for the concentration of iodine gas at equilibrium. Meaning we're going to need to multiply both sides by our denominator. So multiplying by zero point, the concentration of 0.0160 moles per liter for hydrogen and that is multiplied by our concentration of iodine gas. Those terms will then cancel out on the right hand side of our equation. And of course, we are multiplying by the same terms on the left hand side of our equation in our next line. This will simplify so that we have 2.90 times 10 to the negative first power multiplied by our concentration of I'm sorry, this should say 0.0160 no on both sides of our reaction. So 2.90 times 10 to the negative first power is multiplied by our concentration of hydrogen gas, 0.0160 moles per liter, which is multiplied then by our concentration of iodine gas and equal to our concentration of hydrogen iodide gas, 0.0320 moles per liter, which is still being squared. So to isolate for our concentration of iodine gas, we're going to need to isolate it by dividing both sides of our equation. So by dividing both sides of our equation by 2.90 times 10 to the negative first power, so that it will cancel out on the left, which is multiplied by our concentration of hydrogen gas, 0.0160 moles per liter. So that will also cancel out on the left. And again, we are dividing by those terms on the right hand side of our equation as well. Now, on the right hand side of our equation notice that we can cancel out our units of moles per liter. And actually for more accuracy, we have square units of moles per liter in the numerator. So we would have one unit of that remaining for that unit. And thus, in our next line, we would simplify so that our concentration of iodine gas is equal to the proportion between 0.01 or sorry, 0.001 21024, which has units of moles per liter and is divided by the result of our denominator being 4.64 times 10 to the negative third power. And this quotient will simplify so that our concentration of iodine gas at equilibrium is equal to 0.2 to one moles per liter. And this would be our final answer to complete this example as the concentration of iodine gas at equilibrium, given our concentration of our reactants at equilibrium at 300 Kelvin. I hope that this made sense and let us know if you have any questions.

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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

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