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Chem 1310: Introduction to physical chemistry Part 3: Equilibria Using ICE for equilibria
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Write out the chemical equation Setup the ICE table: one column for each reactant and product one row each for Initial/Change/Equilibrium Define one change as x, express the other changes in it, fill the table Use stoichiometry or the rate constant expression to get an equation in x, solve it, calculate all fields
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H 2 + I 2 ⇋ 2 HI stoichiometry question Given initial [H 2 ] = [I 2 ] = 0.0175 M, final [HI] = 0.0270 M, calculate K. [H 2 ][I 2 ][HI] I C E
![H 2 + I 2 ⇋ 2 HI stoichiometry question Given initial [H 2 ] = [I 2 ] = M, final [HI] = M, calculate K.](http://images.slideplayer.com./15/4812744/slides/slide_3.jpg "[H 2 ][I 2 ][HI] I C E.")
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H 2 + I 2 ⇋ 2 HI stoichiometry question Given initial [H 2 ] = [I 2 ] = 0.0175 M, final [HI] = 0.0270 M, calculate K. [H 2 ][I 2 ][HI] I.0175 0 C E
![H 2 + I 2 ⇋ 2 HI stoichiometry question Given initial [H 2 ] = [I 2 ] = M, final [HI] = M, calculate K.](http://images.slideplayer.com./15/4812744/slides/slide_4.jpg "[H 2 ][I 2 ][HI] I C E.")
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H 2 + I 2 ⇋ 2 HI stoichiometry question Given initial [H 2 ] = [I 2 ] = 0.0175 M, final [HI] = 0.0270 M, calculate K. [H 2 ][I 2 ][HI] I.0175 0 C -x-x-x-x+2x E
![H 2 + I 2 ⇋ 2 HI stoichiometry question Given initial [H 2 ] = [I 2 ] = M, final [HI] = M, calculate K.](http://images.slideplayer.com./15/4812744/slides/slide_5.jpg "[H 2 ][I 2 ][HI] I C -x-x-x-x+2x E.")
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H 2 + I 2 ⇋ 2 HI stoichiometry question Given initial [H 2 ] = [I 2 ] = 0.0175 M, final [HI] = 0.0270 M, calculate K. [H 2 ][I 2 ][HI] I.0175 0 C -x-x-x-x+2x E.0175-x 2x2x
![H 2 + I 2 ⇋ 2 HI stoichiometry question Given initial [H 2 ] = [I 2 ] = M, final [HI] = M, calculate K.](http://images.slideplayer.com./15/4812744/slides/slide_6.jpg "[H 2 ][I 2 ][HI] I C -x-x-x-x+2x E.0175-x 2x2x.")
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H 2 + I 2 ⇋ 2 HI stoichiometry question Given initial [H 2 ] = [I 2 ] = 0.0175 M, final [HI] = 0.0270 M, calculate K. [H 2 ][I 2 ][HI] I.0175 0 C -x-x-x-x+2x E.0175-x 2x =.0276
![H 2 + I 2 ⇋ 2 HI stoichiometry question Given initial [H 2 ] = [I 2 ] = M, final [HI] = M, calculate K.](http://images.slideplayer.com./15/4812744/slides/slide_7.jpg "[H 2 ][I 2 ][HI] I C -x-x-x-x+2x E.0175-x 2x =")
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H 2 + I 2 ⇋ 2 HI stoichiometry question Given initial [H 2 ] = [I 2 ] = 0.0175 M, final [HI] = 0.0270 M, calculate K. [H 2 ][I 2 ][HI] I.0175 0 C -x-x-x-x+2x E.0175-x =.0037 2x =.0276 ( x =.0138)
![H 2 + I 2 ⇋ 2 HI stoichiometry question Given initial [H 2 ] = [I 2 ] = M, final [HI] = M, calculate K.](http://images.slideplayer.com./15/4812744/slides/slide_8.jpg "[H 2 ][I 2 ][HI] I C -x-x-x-x+2x E.0175-x = x =.0276 ( x =.0138).")
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H 2 + I 2 ⇋ 2 HI stoichiometry question K = [HI] 2 /([H 2 ][I 2 ]) = (0.0276/0.0037) 2 = 56 [H 2 ][I 2 ][HI] I.0175 0 C -x-x-x-x+2x E.0175-x =.0037 2x =.0276 ( x =.0138)
![H 2 + I 2 ⇋ 2 HI stoichiometry question K = [HI] 2 /([H 2 ][I 2 ]) = (0.0276/0.0037) 2 = 56 [H 2 ][I 2 ][HI] I C -x-x-x-x+2x E.0175-x = x =.0276 ( x =.0138)](http://images.slideplayer.com./15/4812744/slides/slide_9.jpg "H 2 + I 2 ⇋ 2 HI stoichiometry question K = [HI] 2 /([H 2 ][I 2 ]) = (0.0276/0.0037) 2 = 56 [H 2 ][I 2 ][HI] I C -x-x-x-x+2x E.0175-x = x =.0276 ( x =.0138)")
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H 2 + I 2 ⇋ 2 HI initial composition question Given initial [H 2 ] = [I 2 ] = 0.01 M, K = 56, what is the final composition? [H 2 ][I 2 ][HI] I 0.01 0 C E
![H 2 + I 2 ⇋ 2 HI initial composition question Given initial [H 2 ] = [I 2 ] = 0.01 M, K = 56, what is the final composition.](http://images.slideplayer.com./15/4812744/slides/slide_10.jpg "[H 2 ][I 2 ][HI] I C E.")
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H 2 + I 2 ⇋ 2 HI initial composition question Given initial [H 2 ] = [I 2 ] = 0.01 M, K = 56, what is the final composition? [H 2 ][I 2 ][HI] I 0.01 0 C -x +2x E
![H 2 + I 2 ⇋ 2 HI initial composition question Given initial [H 2 ] = [I 2 ] = 0.01 M, K = 56, what is the final composition.](http://images.slideplayer.com./15/4812744/slides/slide_11.jpg "[H 2 ][I 2 ][HI] I C -x +2x E.")
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H 2 + I 2 ⇋ 2 HI initial composition question Given initial [H 2 ] = [I 2 ] = 0.01 M, K = 56, what is the final composition? [H 2 ][I 2 ][HI] I 0.01 0 C -x +2x E 0.01-x 2x2x
![H 2 + I 2 ⇋ 2 HI initial composition question Given initial [H 2 ] = [I 2 ] = 0.01 M, K = 56, what is the final composition.](http://images.slideplayer.com./15/4812744/slides/slide_12.jpg "[H 2 ][I 2 ][HI] I C -x +2x E 0.01-x 2x2x.")
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H 2 + I 2 ⇋ 2 HI initial composition question Use K = [HI] 2 /([H 2 ][I 2 ]) = (2x) 2 /(0.01-x) 2 = 56 ("trick") (2x)/(0.01-x) = 7.48 2x = 0.0748-7.48x x = 0.00789 [H 2 ][I 2 ][HI] I 0.01 0 C -x +2x E 0.01-x 2x2x
![H 2 + I 2 ⇋ 2 HI initial composition question Use K = [HI] 2 /([H 2 ][I 2 ]) = (2x) 2 /(0.01-x) 2 = 56 ( trick ) (2x)/(0.01-x) = 7.48 2x = x x = [H 2 ][I 2 ][HI] I C -x +2x E 0.01-x 2x2x](http://images.slideplayer.com./15/4812744/slides/slide_13.jpg "H 2 + I 2 ⇋ 2 HI initial composition question Use K = [HI] 2 /([H 2 ][I 2 ]) = (2x) 2 /(0.01-x) 2 = 56 ( trick ) (2x)/(0.01-x) = 7.48 2x = x x = [H 2 ][I 2 ][HI] I C -x +2x E 0.01-x 2x2x")
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H 2 + I 2 ⇋ 2 HI initial composition question x = 0.00789 [H 2 ][I 2 ][HI] I 0.01 0 C -x +2x E 0.01-x = 0.0021 2x = 0.0158
![H 2 + I 2 ⇋ 2 HI initial composition question x = [H 2 ][I 2 ][HI] I C -x +2x E 0.01-x = x =](http://images.slideplayer.com./15/4812744/slides/slide_14.jpg "H 2 + I 2 ⇋ 2 HI initial composition question x = [H 2 ][I 2 ][HI] I C -x +2x E 0.01-x = x =")
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H 2 + I 2 ⇋ 2 HI initial composition question 2 Given initial [H 2 ] = 0.01 M, [I 2 ] = 0.02 M, K = 56, what is the final composition? [H 2 ][I 2 ][HI] I 0.010.020 C E
![H 2 + I 2 ⇋ 2 HI initial composition question 2 Given initial [H 2 ] = 0.01 M, [I 2 ] = 0.02 M, K = 56, what is the final composition.](http://images.slideplayer.com./15/4812744/slides/slide_15.jpg "[H 2 ][I 2 ][HI] I C E.")
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H 2 + I 2 ⇋ 2 HI initial composition question 2 Given initial [H 2 ] = 0.01 M, [I 2 ] = 0.02 M, K = 56, what is the final composition? [H 2 ][I 2 ][HI] I 0.010.020 C -x +2x E
![H 2 + I 2 ⇋ 2 HI initial composition question 2 Given initial [H 2 ] = 0.01 M, [I 2 ] = 0.02 M, K = 56, what is the final composition.](http://images.slideplayer.com./15/4812744/slides/slide_16.jpg "[H 2 ][I 2 ][HI] I C -x +2x E.")
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H 2 + I 2 ⇋ 2 HI initial composition question 2 Given initial [H 2 ] = 0.01 M, [I 2 ] = 0.02 M, K = 56, what is the final composition? [H 2 ][I 2 ][HI] I 0.010.020 C -x +2x E 0.01-x0.02-x2x2x
![H 2 + I 2 ⇋ 2 HI initial composition question 2 Given initial [H 2 ] = 0.01 M, [I 2 ] = 0.02 M, K = 56, what is the final composition.](http://images.slideplayer.com./15/4812744/slides/slide_17.jpg "[H 2 ][I 2 ][HI] I C -x +2x E 0.01-x0.02-x2x2x.")
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H 2 + I 2 ⇋ 2 HI initial composition question 2 K = (2x) 2 /[(0.01-x)(0.02-x)] = 56 4x 2 = 56(0.01-x)(0.02-x) = 56x 2 -1.68x+0.0112 52x 2 -1.68x+0.0112 = 0 x = 0.0094 [H 2 ][I 2 ][HI] I 0.010.020 C -x +2x E 0.01-x0.02-x2x2x
![H 2 + I 2 ⇋ 2 HI initial composition question 2 K = (2x) 2 /[(0.01-x)(0.02-x)] = 56 4x 2 = 56(0.01-x)(0.02-x) = 56x x 52x x = 0 x = [H 2 ][I 2 ][HI] I C -x +2x E 0.01-x0.02-x2x2x](http://images.slideplayer.com./15/4812744/slides/slide_18.jpg "H 2 + I 2 ⇋ 2 HI initial composition question 2 K = (2x) 2 /[(0.01-x)(0.02-x)] = 56 4x 2 = 56(0.01-x)(0.02-x) = 56x x 52x x = 0 x = [H 2 ][I 2 ][HI] I C -x +2x E 0.01-x0.02-x2x2x")
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H 2 + I 2 ⇋ 2 HI initial composition question 2 x = 0.0094 [H 2 ][I 2 ][HI] I 0.010.020 C -x +2x E 0.01-x = 0.0006 0.02-x = 0.0106 2x = 0.0188
![H 2 + I 2 ⇋ 2 HI initial composition question 2 x = [H 2 ][I 2 ][HI] I C -x +2x E 0.01-x = x = x =](http://images.slideplayer.com./15/4812744/slides/slide_19.jpg "H 2 + I 2 ⇋ 2 HI initial composition question 2 x = [H 2 ][I 2 ][HI] I C -x +2x E 0.01-x = x = x =")
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H 2 + I 2 ⇋ 2 HI mixed composition question Given initial [H 2 ] = 0.01 M, K = 56, which initial [I 2 ] should we choose to arrive at a final [HI] = 0.015? [H 2 ][I 2 ][HI] I 0.01y0 C E
![H 2 + I 2 ⇋ 2 HI mixed composition question Given initial [H 2 ] = 0.01 M, K = 56, which initial [I 2 ] should we choose to arrive at a final [HI] =](http://images.slideplayer.com./15/4812744/slides/slide_20.jpg "[H 2 ][I 2 ][HI] I 0.01y0 C E.")
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H 2 + I 2 ⇋ 2 HI mixed composition question Given initial [H 2 ] = 0.01 M, K = 56, which initial [I 2 ] should we choose to arrive at a final [HI] = 0.015? [H 2 ][I 2 ][HI] I 0.01y0 C -x +2x E
![H 2 + I 2 ⇋ 2 HI mixed composition question Given initial [H 2 ] = 0.01 M, K = 56, which initial [I 2 ] should we choose to arrive at a final [HI] =](http://images.slideplayer.com./15/4812744/slides/slide_21.jpg "[H 2 ][I 2 ][HI] I 0.01y0 C -x +2x E.")
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H 2 + I 2 ⇋ 2 HI mixed composition question Given initial [H 2 ] = 0.01 M, K = 56, which initial [I 2 ] should we choose to arrive at a final [HI] = 0.015? [H 2 ][I 2 ][HI] I 0.01y0 C -x +2x E 0.01-xy-xy-x2x = 0.015 ( x = 0.0075)
![H 2 + I 2 ⇋ 2 HI mixed composition question Given initial [H 2 ] = 0.01 M, K = 56, which initial [I 2 ] should we choose to arrive at a final [HI] =](http://images.slideplayer.com./15/4812744/slides/slide_22.jpg "[H 2 ][I 2 ][HI] I 0.01y0 C -x +2x E 0.01-xy-xy-x2x = ( x = ).")
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H 2 + I 2 ⇋ 2 HI mixed composition question K = (0.015) 2 /(0.0025)(y-0.0075) = 56 y = 0.0091 final [I 2 ] = 0.0016 Check by re-calculating K !!! [H 2 ][I 2 ][HI] I 0.01y0 C -x +2x E 0.01-x = 0.0025 y-x = y-0.0075 2x = 0.015 ( x = 0.0075)
![H 2 + I 2 ⇋ 2 HI mixed composition question K = (0.015) 2 /(0.0025)(y ) = 56 y = final [I 2 ] = Check by re-calculating K !!.](http://images.slideplayer.com./15/4812744/slides/slide_23.jpg "[H 2 ][I 2 ][HI] I 0.01y0 C -x +2x E 0.01-x = y-x = y x = ( x = ).")
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Ammonia formation Given a mixture of 1 mol N 2 and 3 mol H 2 in 10L, we add a catalyst to form ammonia at room temperature. What percentage of the starting material will react? K C = 3.5*10 8. N 2 + 3 H 2 ⇋ 2 NH 3
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Ammonia formation [N 2 ][H 2 ][NH 3 ] I 0.10.30 C -x-x-3x+2x E 0.1-x0.3-3x2x2x 1 mol N 2 and 3 mol H 2 in 10L
![Ammonia formation [N 2 ][H 2 ][NH 3 ] I C -x-x-3x+2x E 0.1-x0.3-3x2x2x 1 mol N 2 and 3 mol H 2 in 10L](http://images.slideplayer.com./15/4812744/slides/slide_25.jpg "Ammonia formation [N 2 ][H 2 ][NH 3 ] I C -x-x-3x+2x E 0.1-x0.3-3x2x2x 1 mol N 2 and 3 mol H 2 in 10L")
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Ammonia formation K C = [NH 3 ] 2 /([N 2 ][H 2 ] 3 ) = (2x) 2 /[(.1-x)(.3-3x) 3 ] = 4x 2 /[27(.1-x) 4 ] = 3.5*10 8 2x/[5.196(.1-x) 2 ] = 18708 (.1-x) 2 = 0.0000206x x 2 -0.20002x+0.01 = 0 x = 0.0986 [N 2 ][H 2 ][NH 3 ] I 0.10.30 C -x-x-3x+2x E 0.1-x0.3-3x2x2x
![Ammonia formation K C = [NH 3 ] 2 /([N 2 ][H 2 ] 3 ) = (2x) 2 /[(.1-x)(.3-3x) 3 ] = 4x 2 /[27(.1-x) 4 ] = 3.5*10 8 2x/[5.196(.1-x) 2 ] = (.1-x) 2 = x x x+0.01 = 0 x = [N 2 ][H 2 ][NH 3 ] I C -x-x-3x+2x E 0.1-x0.3-3x2x2x](http://images.slideplayer.com./15/4812744/slides/slide_26.jpg "Ammonia formation K C = [NH 3 ] 2 /([N 2 ][H 2 ] 3 ) = (2x) 2 /[(.1-x)(.3-3x) 3 ] = 4x 2 /[27(.1-x) 4 ] = 3.5*10 8 2x/[5.196(.1-x) 2 ] = (.1-x) 2 = x x x+0.01 = 0 x = [N 2 ][H 2 ][NH 3 ] I C -x-x-3x+2x E 0.1-x0.3-3x2x2x")
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Ammonia formation x = 0.0986. Conversion percentage: (x/0.1)*100 = 98.6% Try this for yourself: what happens if the volume is increased to 100 L? [N 2 ][H 2 ][NH 3 ] I 0.10.30 C -x-x-3x+2x E 0.1-x =.0014 0.3-3x =.0042 2x =.0028
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Formation of water - using approximations O 2 + 2 H 2 ⇋ 2 H 2 O We start with 1 mol of O 2, 2 mol of H 2 in 10 L. How many molecules of reactants remain?
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Formation of water - using approximations K C = [H 2 O] 2 /([O 2 ][H 2 ] 2 ) = (2x) 2 /[(.1-x)(.2-2x) 2 ] = (2x) 2 /[4(.1-x) 3 ] = 3.3*10 81. Not easy to solve. Problem! [O 2 ][H 2 ][H 2 O] I 0.10.20 C -x-x-2x+2x E 0.1-x0.2-2x2x2x
![Formation of water - using approximations K C = [H 2 O] 2 /([O 2 ][H 2 ] 2 ) = (2x) 2 /[(.1-x)(.2-2x) 2 ] = (2x) 2 /[4(.1-x) 3 ] = 3.3*10 81.](http://images.slideplayer.com./15/4812744/slides/slide_29.jpg "Not easy to solve. Problem. [O 2 ][H 2 ][H 2 O] I C -x-x-2x+2x E 0.1-x0.2-2x2x2x.")
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Formation of water - using approximations K C = (2x) 2 /[4(.1-x) 3 ] = 3.3*10 81 x must be very close to 0.1. There is a standard "trick" to solve problems like this. You want to neglect a small variable relative to a larger known quantity. That works best if we are trying to solve for a small variable. Since the reaction is virtually complete, it is better to solve for the fraction (y) that has not reacted.
![Formation of water - using approximations K C = (2x) 2 /[4(.1-x) 3 ] = 3.3*10 81 x must be very close to 0.1.](http://images.slideplayer.com./15/4812744/slides/slide_30.jpg "There is a standard trick to solve problems like this. You want to neglect a small variable relative to a larger known quantity. That works best if we are trying to solve for a small variable. Since the reaction is virtually complete, it is better to solve for the fraction (y) that has not reacted..")
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K C = [H 2 O] 2 /([O 2 ][H 2 ] 2 ) = (0.2-2y) 2 /[y(2y) 2 ] = (0.2-2y) 2 /[4y 3 ] (0.2) 2 /[4y 3 ] = 3.3*10 81 y 3 = (0.2) 2 /(4*3.3*10 81 ) = 3.03*10 -84 y = 1.45*10 -28 mol/L or a total of 0.00087 molecules. "Product-favoured" Formation of water - using approximations [O 2 ][H 2 ][H 2 O] I 000.2 C +y+y+2y-2y E y2y2y0.2-2y
![K C = [H 2 O] 2 /([O 2 ][H 2 ] 2 ) = (0.2-2y) 2 /[y(2y) 2 ] = (0.2-2y) 2 /[4y 3 ] (0.2) 2 /[4y 3 ] = 3.3*10 81 y 3 = (0.2) 2 /(4*3.3*10 81 ) = 3.03* y = 1.45* mol/L or a total of molecules.](http://images.slideplayer.com./15/4812744/slides/slide_31.jpg "Product-favoured Formation of water - using approximations [O 2 ][H 2 ][H 2 O] I C +y+y+2y-2y E y2y2y0.2-2y.")
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