1 / 7

Equilibrium Chemical Reactions and Calculations

Learn how to calculate equilibrium concentrations and constants for chemical reactions at varying temperatures.

prusso
Download Presentation

Equilibrium Chemical Reactions and Calculations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Hydrogen iodide decomposes at moderate temperatures according to the equation 2HI(g) <----> H2(g) + I2(g) The amount of I2 in the mixture can be determined from the intensity of the violet color of I2; the more intense, the more I2 in the reaction vessel. When 4.00 mol of HI were placed in a 5.00 L vessel at 458°C, the equilibrium mixture was found to contain 0.442 mol I2. What is the value of K? Initial: 4.00 mol 0.0 mol 0.0 mol At Eq: ? ? 0.442 mol

  2. Initial: 4.00 mol 0.0 mol 0.0 mol 2HI(g) <----> H2(g) + I2(g) At Eq.: ? ? 0.442 mol 0.442 mol Which reaction proceeds initially? Why? The forward…no products to make reverse go at first! 0.442 mol of I2 produced…how much H2? 0.442 mol of H2…one to one molar ratio and they both started out the same, zero

  3. Initial: 4.00 mol 0.0 mol 0.0 mol 2HI(g) <----> H2(g) + I2(g) At Eq.: ? ? 0.442 mol 0.442 mol Where did the I2 and the H2 come from? The HI, obviously! How much of that was used? 0.884 mol of HI…two to one molar ratio

  4. Initial: 4.00 mol 0.0 mol 0.0 mol 2HI(g) <----> H2(g) + I2(g) -0.884 mol At Eq.: ? ? 0.442 mol 3.12 mol 0.442 mol If we started out with 4.00 mol of HI initially, how much of it remains at equilibrium then? 3.12 mol of HI at eq. The concentration of HI at equilibrium would be? (Remember the reaction vessel was 5.00 L) 0.624 M of HI

  5. Initial: 4.00 mol 0.0 mol 0.0 mol 2HI(g) <----> H2(g) + I2(g) At Eq.: ? ? 0.442 mol 3.12 mol 0.442 mol (0.0884 M) (0.624 M) (0.0884 M) The [H2] = [I2] = ? 0.0884 M = [H2] = [I2] Now, determine the K value... K = (0.0884M) (0.0884M)/ (0.624M)2 = 0.0201

  6. For the reaction 2 IBr(g) <----> I2(g) + Br2(g) K = 2.5 x 10-3 at 315 K Calculate the equilibrium concentration of each species in a 4.0 L vessel starting with 0.30 mol of each substance. Also, what is the value of Kp. IBr Initial: 0.075 M 0. 075 M 0.075 M 4.0 L +2x -x -x Change: At Eq: 0.075M + 2x ? 0.075M - x ? 0.075M - x ? I2 Br2 0.21M 0.010M 0.010M (0.075M)(0.075M) = Q reaction quotient = 1 Kp= (0.075M)2 K= (0.075-x)(0.075-x) K= (0.075-x)2 Q>K (the reverse reaction is favored) = 2.5 x 10-3 (0.075+2x)2 0.075-x = 0.050 x = 0.065 0.075+2x

  7. Consider the following reaction that takes place at 773 K CO(g) +H2O(g) <----> CO2(g) + H2(g) The reaction took place by injecting into a reaction vessel the materials shown below to produce the indicated initial conditions. carbon monoxide: 0.0300 M water: 0.0400 M carbon dioxide: 0.0800M hydrogen: 0.0300 M Find the equilibrium concentrations of all the substances given the equilibrium constant, Kc, is equal to 4.00 at this temperature. x = 0.00648 students who answer correctly before the bell rings will receive 2 bonus points! (no sharing of info!) Equilibrium Concentrations: [CO] = 0.235 M [H2O]= 0.0335 [CO2]= 0.865 M [H2] = 0. 365 M

More Related