Hydrogen iodide decomposes according to the reaction 2 HI(g) ⇌ H2(g) + I2(g) A sealed 1.50−L container initially holds 0.00623 mol of H2, 0.00414 mol of I2, and 0.0244 mol of HI at 703 K. When equilibrium is reached, the concentration of H2(g) is 0.00467 M. What are the concentrations of HI(g) and I2(g)?

Answered: Image /qna-images/answer/4b760176-9b56-412a-92bc-ff84a4f59599.jpgNow 2HI ----------> H2 + I2 Initial concentration 0.01627 0.004153 0.00276 Change -2x +x +x Equilibrium .01627-2x 0.004153+x .00276+x Concentration Now equilibria concentration of H2 is 0.00467M so at equilibrium concentration of H2 is as 0.004153 + x = 0.00467 M So x = 0.000517 So concentration of HI at equilibrium is 0.01627-2x = 0.01627-2*0.000517 = 0.01523M Similarly concentration of I2= 0.00276+x = 0.00276 + 0.000517 = 0.003277 M

Answered: A sealed 1.50−L container initially…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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Hydrogen iodide decomposes according to the reaction

2 HI(g) ⇌ H₂(g) + I₂(g)

A sealed 1.50−L container initially holds 0.00623 mol of H₂, 0.00414 mol of I₂, and 0.0244 mol of HI at 703 K. When equilibrium is reached, the concentration of H₂(g) is 0.00467 M. What are the concentrations of HI(g) and I₂(g)?

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