Textbook 5I.27
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Textbook 5I.27
I got part A) where Q=0.50 and so it is not at equilibrium as compared to Kc=0.56. Since Q<K that means for part b) the reaction will shift to the right towards products. For c) will I have to use the quadratic formula? Also Can someone confirm my answers are correct for this textbook problem. Thanks so much, anything helps.
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Re: Textbook 5I.27
Your answers for a and b are correct, and part c does require the quadratic formula. Since we don't know x and K is far too large for us to approximate equilibrium molarity of PCl3, we must find x by rearranging the K equation into the ax^2 + bx + c = 0 format and solve with the formula. In this problem, the K equation before making any steps to solve it is 0.56 = (3+x)/[(6-x)(1-x)]
Re: Textbook 5I.27
Your answers for part A and B are correct, and yes for C you will have to use the quadratic equation to solve for x. When you rearrange the chemical equilibrium equation so that it matches after you've plugged in the equilibrium concentrations, you should get the standard equation: 0.56x^2 -4.92x + 0.36 = 0. Solving for x afterwards is just plugging the values into your calculator using the quadratic.
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