The reaction of fumarate with water to form L-malate is catalyzed by the enzyme fumarase; Kc = 3.3 at 37°C. When a reaction mixture with [fumarate] = 1.56 * 10-3 M and [l -malate] = 2.27 * 10-3 M comes to equilibrium in the presence of fumarase at 37 °C, what are the equilibrium concentrations of fumarate and L-malate? (Water can be omit- ted from the equilibrium equation because its concentration in dilute solutions is essentially the same as that in pure water.)

Video transcript

Hello. And this problem we're told the enzyme nitrile hydra taste catalyze is the reaction of Melanie trials with water to form Siano car box. Um Eyes with an equilibrium costs of 4.89 at 25 degrees Celsius. We are asked what are the equilibrium concentrations of our react in some products? In the reaction mixture has a mild night trials concentration of 0.135. And the cyanide carb oxidized concentration is 0.128 moller reaches equilibrium in the presence of nitrile hydro test at 25 degrees Celsius. Due to the fact that water's concentration is diluted solutions is nearly identical to that of pure water. It could be excluded from the equilibrium equation. So we'll begin by creating an ice table. So we have our reactant going to form our products and we have our initial amount an initial amount of our reactant 0.135 moller. And our product is 0.128 Mueller. The change then is minus X. And then plus X. And equilibrium amount is our initial interchange combined. So 0.135 minus X. And 0.128 plus X. Writing in the equilibrium constant expression we get K. C. Then is equal to the concentration of our products over the concentration of our reactant Withstand plug in values from the ice table is 0.00128 plus x. Over 0.00135 -1. And we're told that is equal to 4.89. So then we have 0.128 plus x is equal to 4.89 times 0.135 minus X. On the right hand side. This works out to then 0. -4.89 x. Moving our X values to one side and those without an X to the other side, We get x plus 4.89 x Is equal 0.0066, : -0.00128. We have now 5.89, x is equal to zero point 0053, So solving for XX is now equal to 0.0053, divided by 5.89, Which is equal to 9.03 times 10 to the -4. So at equilibrium then our concentration of reactant Looking back at the ice table is equal to 0.0013, -1. To be equal to 0.0013, -9.03 times 10 : -4. So this works out to 4.47. I'm sent to them -4 Mueller. And then our concentration of products Looking at the ice table is given by 0.001, 2, 8 plus x. This works out to 0.128 plus 9.3 times 10 to the minus four, Which is equal to 0.00218 Mueller. So our reactant concentration at equilibrium is equal to 4.47 times 10 to minus four Mueller, and our product concentration is equal to 0.218 Mueller at equilibrium. Thanks for watching. Hope. This helps.