I am able to do the \"Example\" problem and get the correct answer. But I am una
ID: 545069 • Letter: I
Question
I am able to do the "Example" problem and get the correct answer. But I am unable to get the correct answer for the "Practice problem" circled in red. I do not understand how to set up the ICE table (with the green arrow) for the equation, in order to receive the proper answer
Example 14.4.4 Using a Reactlon Table to Calculate K The gas OF (g) has been studied as a possible oxidizing agent for rocket fuels. Calculate the value of Kp at 25 C for the production of OF,(g) from the elements if oxygen gas at 0.200 atm and fluorine gas at 0.300 atm are allowed to react, and at equilibrium the pressure of gas is 0.100 atm. The equation is O(g)+2F,8) -20F, Solution We first construct the reaction table, filling in the information given in the problem. It is implied in the statement of the problem that there is no OF,(g) present initially 0.200 atm 0.300 atmm 0 atm I 0.100 atm From the table we can see that the pressure of O, has dropped from 0.200 atm to 0.100 atm, a decrease of 0.100 atm. Because of the 2:1 stoichiometric ratio, the F, pressure will drop twice as much as the O, presure, or 0.200 atm, and the OF, presure will ncrease by 0.200 am This allows us to complete the table. 2F2(g) 0.300 atm -20F2(g) 0 atm 0.200 atm -0.100 atm 2(0.100 atm) +2(0.100 atm) C 0.100 atm 0.200 atm 0.100 atm Pressure changes during a reaction must reflect the stoichiometric ratios of the balanced equa tion, just as concentration changes do. The equilibrium pressure in each column is the alge- braic sum of the Initial and Change values. We can now calculate the value of the equilibrium constant. (PoF) (0.200 atm)2 40.0 atm -1 F) (0.100 atm ractice Problem 14.4.4 Ithough Kp for the formation of NH,(g) from its elements is large at room temperature, hydrogen gas at 0.410 atm are combined, and after equilibrium is established the pressure o decreases as the temperature increases. Calculate Kp at 227 "C if nitrogen gas at 0.220 atm H,(g) is only 0.040 atm. The equation is N,(8+3H(2NH,().Explanation / Answer
N2. + 3H2 2NH3
Initial 0.220. O.410. 0
Equilibrium 0.220- 0.040 0.410- 3(0.040) 0.040
0.18. 0.29 0.040
Kp. = [NH3]^2/ [N2] [H2]^3
= (0.040)^2 / (0.18) (0.29)^3
= 0.36. atm -2
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