Jump to... al Chemistry 4th Edition University Science Books presented by Saplin

Question

Jump to... al Chemistry 4th Edition University Science Books presented by Sapling Learning Ma At 500 "C hydrogen iodide decomposes according to 2HI For HIl(g) heated to 500 "C in a 1.00-L reaction vessel, chemical analysis determined the following concentrations at equilibrium: [H2l0.324 M, [l 0.324 M, and [HI] 2.74 M. If an additional 1.00 mole of HI(g) is introduced into the reaction vessel, what are the equilibrium concentrations after the new equilibrium has been reached? Number HI] Number Number [12]- Previous Give Up & View Solution Check AnswerNext Exit int about us careers privacy policy terms of use contact us help

Explanation / Answer

Given

2 HI <---> H2 (g) + I2 (g)

first equilibrium

[I2] = 0.324 M

[H2] = 0.324 M

[HI] = 2.74 M

k = [I2][H2] / [HI]2 = 0.324 * 0.324 / (2.74)2 = 0.014

Given

Volume = 1 L

[HI] = 2.74 mol/L

no. of moles of HI = 2.74 mol/L * 1 L = 2.74 moles

second equilibrium

1 mole of HI is added

so no. of moles of HI = 2.74 + 1 = 3.74 moles

volume = 1 L

[HI] = 3.74 mol / 1 L = 3.74 mol/L

2 HI <--> H2 + I2

initial 3.74 0.324 0.324

convert -x x x

final 3.74-x 0.324+x 0.324+x

k = 0.014 = (0.324 + x ) ( 0.324 + x) / (3.74 - x)2 = 0.105 + 0.648 x + x2 / ( 13.9876 - 7.48 x + x2 )

0.196 - 0.105 x + 0.014 x2 = 0.105 + 0.648 x + x2  

0.986 x2 + 0.753 x - 0.091 = 0

x = 0.106

[HI] = 3.74 - x = 3.74 - 0.106 = 3.634 M

[H2] = 0.324 + 0.106 = 0.43 M

[I2] = 0.324 + 0.106 = 0.43 M

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