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0 saplinglearning.com O G a Dashboard 5cos(21)=y"+4y - Laplace Tran... University of Kentucky - CHE 1... 5cos(21)y'+4y-Laplace Transfo... Dashboard 5cos2t-y"44y-Lapl... O Saplinglearning.com O Do D. + a Sapling Learning Jump to... macmillan learning View Profile Logout Sapling Learning > University of Kentucky - CHE 107 - Spring18-DEPT > Activities and Due Dates > HW 412 My Assignment Resources # Attempts Score O Gradebook O Assignment Information 0 3/30/2018 11:59 PM $ 56.6/100 e Print I Calculator - Periodic Table Question 9 of 15 1 2 98 2 1 100 Map. M General Chemistry 4th Edition ate Rook Gallo University Science Books presented by Sapling Leaming 3 1 100 At 500 °C hydrogen iodide decomposes according to Available From: Not Set Due Date: 3/30/2018 11:59 PM Points Possible: 100 Grade Category: Homework Description: Policies: Homework 100 2HI(g) 2 H, g) +1,8) 5 2 95 For HI(g) heated to 500 °C in a 1.00-L reaction vessel, chemical analysis determined the following concentrations at equilibrium: [H] = 0.425 M, [12] = 0.425 M, and [HI] = 3.59 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? 60 7 0 0 Number You can check your answers, You can view solutions when you complete or give up on any question. You can keep trying to answer each question until you get it right or give up. You lose 5% of the points available to each answer in your question for each incorrect attempt at that answer. [HI] = 9 0 0 10 1 100 OeTextbook [H.] = 11 0 Number O Help With This Topic 0 12 13 0 2 1,] = o Web Help & Videos 14 2 98 O Technical Support and Bug Reports 15463 O Previous & Give Up &. View Solution Check Answer ONext ] Exit

Explanation / Answer

For the reaction,

K = [H2][I2]/[HI]^2

Taking equilibrium concentrations,

K = (0.425)(0.425)/(3.59)^2 = 0.014

when addition 1.0 mole HI is added

new [HI] = (3.59 M x 1.0 L + 1.0 moles)/1.0 L = 4.59 M

ICE chart

        2HI(g) <==> H2(g)   + I2(g)

I        4.59            0.425        0.425

C       -2x               +x             +x

E     4.59-2x       0.425+x    0.425+x

So,

0.014 = (0.425+x)(0.425+x)/(4.59-2x)^2

0.944x^2 + 1.11x - 0.12 = 0

x = 0.1 M

So neq equilibrium concentrations of,

[HI] = 4.59 - 2 x 0.1 = 4.39 M

[H2] = 0.425 + 0.1 = 0.525 M

[I2] = 0.425 + 0.1 = 0.525 M

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