Recalculate the equilibrium concentrations in Problem 15.93 if th...

Question

Recalculate the equilibrium concentrations in Problem 15.93 if the initial concentrations are 2.24 M N2 and 0.56 M O2. (This N2>O2 concentration ratio is the ratio found in air.)

Accepted Answer

Hello. In this problem, we are told hydrogen iodide with an initial concentration of 0.5 Mueller dissociates into hydrogen and iodine gas. Were asked to calculate the equilibrium concentration of hydrogen iodide hydrogen and iodine in the reaction mixture at 300 kelvin. To determine the equilibrium concentrations, we need to generate an ice table. So we begin by writing our reaction. So we have two moles of hydrogen iodide decomposes to form hydrogen gas and I dine gas. We have initial change and equilibrium. Initially we have 0.500 Mueller of our hydrogen iodide and initially no hydrogen or ii. Our change then is minus two X. Based on the coefficient balanced reaction equation and the fact that hydrogen iodide is being consumed and then we have plus X plus X. Our equilibrium then we combine the initial the change so that 20.500 minus two X. X. And X. Well now right. Our equilibrium constant expression which is equal to the concentration of hydrogen times the concentration of iodine all over the concentration of hydrogen iodide squared. Making use of values from the ice table. This works out to X times X. All over 0.500 minus two X. And this is squared. This then works out to X squared over 0.500 minus two X squared. We're told that this is equal to 3.50 Times 10 -3. So we can simplify this by taking the square root of both sides. In doing so, we then get X about two x 0.500 - x equal to the square root of 3.50 times 10 to minus three, Which is equal to 0.05916. So we move our denominator over to the right hand side. You get X. is equal to 0.05916, 0.5, -2 x. It simplifies to 0.0295, eight -0.1183 x. Move all of our values with variable X. one side, You get 1.1183 x Is equal to 0.02958. So solving for X, Get X is equal to 0.2958 by by 1.1183, Which is equal to 2.6, 5 Times 10 : -2. Some of them are ice table. You know, then the X. Is equal to the concentration of hydrogen And the concentration of iodine at equilibrium, these are both equal to 2.65 times 10 the -2 molar. And our concentration of hydrogen iodide From the ice table at Equilibrium is equal to .500 - x. And so this works out to 0.447 Moller. So at equilibrium are concentrations of hydrogen iodide are 2.65 times 10 minus two and our concentration of hydrogen iodide is 0. Moeller. Thanks for watching. Hope. This helped.

Recalculate the equilibrium concentrations in Problem 15.93 if the initial concentrations are 2.24 M N2 and 0.56 M O2. (This N2>O2 concentration ratio is the ratio found in air.)

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