### Equilibrium Concentration Calculation for the Reaction at 698 KThe equilibrium constant, \( K_c \), for the following reaction is \( 1.80 \times 10^{-2} \) at 698 K.\[ 2\text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \]Calculate the equilibrium concentrations of reactant and products when 0.256 moles of HI are introduced into a 1.00 L vessel at 698 K.- \([\text{HI}] =\) \_\_\_\_ M- \([\text{H}_2] =\) \_\_\_\_ M- \([\text{I}_2] =\) \_\_\_\_ M

Given, moles of HI = 0.256mol Kc = 0.0180

The equilibrium constant, K., for the following reaction is 1.80×10² at 698 K. 2HI(g)=H2(g) + I½(g) Calculate the equilibrium concentrations of reactant and products when 0.256 moles of HI are introduced into a 1.00 L vessel at 698 K. (HI] M [H2] M [12] M

The equilibrium constant, K., for the following reaction is 1.80×10² at 698 K. 2HI(g)=H2(g) + I½(g) Calculate the equilibrium concentrations of reactant and products when 0.256 moles of HI are introduced into a 1.00 L vessel at 698 K. (HI] M [H2] M [12] M

Chemistry

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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

Section: Chapter Questions

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Question

### Equilibrium Concentration Calculation for the Reaction at 698 K

The equilibrium constant, \( K_c \), for the following reaction is \( 1.80 \times 10^{-2} \) at 698 K.

\[ 2\text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \]

Calculate the equilibrium concentrations of reactant and products when 0.256 moles of HI are introduced into a 1.00 L vessel at 698 K.

- \([\text{HI}] =\) \_\_\_\_ M
- \([\text{H}_2] =\) \_\_\_\_ M
- \([\text{I}_2] =\) \_\_\_\_ M

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