Hydrogen iodide decomposes according to the reaction:2 HI(g) = H2(g) + I2(g)At 703 K, a sealed 1.50-L container initially holds0.00415 M H2,0.00276 M I2, and0.0163 M HIWhen equilibrium is reached, the equilibrium concentration of H2(g) is 0.00467 M. What is theequilibrium concentration of HI(g)?None of these0.00328 M0.00052 M0.0158 M0.0173 M0.0153 M

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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)
At 703 K, a sealed 1.50-L container initially holds
0.00415 M H2,
0.00276 M I2, and
0.0163 M HI
When equilibrium is reached, the equilibrium concentration of H2(g) is 0.00467 M. What is the
equilibrium concentration of HI(g)?
None of these
0.00328 M
0.00052 M
0.0158 M
0.0173 M
0.0153 M

Expert Solution

Introduction

In equilibrium concepts, it is quite common to come across various equilibrium constants. The equilibrium constant (K) becomes quite helpful while expressing the relationship between products and reactants for a particular reaction (under equilibrium conditions).

Given data

The temperature is 703 K and the volume is 1.50 L.

The initial concentrations are:

[H₂] = 0.00415 M

[I₂] = 0.00276 M

[HI] = 0.0163 M

The equilibrium concentration of H2 is 0.00467 M.

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