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The equilibrium constant, Kc, for the following reaction is 0.0180 at 698 K. 2HI(g) → H₂(g) + 1₂ (g) Calculate the equilibrium concentrations of reactant and products when 0.293 moles of HI(g) are introduced into a 1.00 L vessel at 698 K. [HI] = = [H₂] [1₂] = M M M

The equilibrium constant, Kc, for the following reaction is 0.0180 at 698 K. 2HI(g) → H₂(g) + 1₂ (g) Calculate the equilibrium concentrations of reactant and products when 0.293 moles of HI(g) are introduced into a 1.00 L vessel at 698 K. [HI] = = [H₂] [1₂] = M M M

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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
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The equilibrium constant, Kc, for the following reaction is 0.0180 at 698 K. 2HI(g) ⇒ H₂(g) + 1₂ (9) Calculate the equilibrium concentrations of reactant and products when 0.293 moles of HI(g) are introduced into a 1.00 L vessel at 698 K. [HI] = [H₂] [1₂] = = ΣΣΣ
Transcribed Image Text:The equilibrium constant, Kc, for the following reaction is 0.0180 at 698 K. 2HI(g) ⇒ H₂(g) + 1₂ (9) Calculate the equilibrium concentrations of reactant and products when 0.293 moles of HI(g) are introduced into a 1.00 L vessel at 698 K. [HI] = [H₂] [1₂] = = ΣΣΣ
Expert Solution
Step 1

Given,

At 698 K,

2HI(g) ⇌ H2(g) + I2(g)

Equilibrium constant (Kc) = 0.0180

Initially moles of HI = 0.293 moles

Volume of the vessel = 1.00 L

Equilibrium concentration of reactant and product,

[HI] = ? M

[H2] = ? M

[I2] = ? M

 

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