We’re being asked to **determine the equilibrium concentration of H _{2}** at 700 K for

H_{2}(g) + I_{2}(g) ⇌ 2 HI(g)

Recall that the ** equilibrium constant** is the ratio of the products and reactants.

We use **K _{p}** when dealing with pressure and

$\overline{){{\mathbf{K}}}_{{\mathbf{p}}}{\mathbf{=}}\frac{{\mathbf{P}}_{\mathbf{products}}}{{\mathbf{P}}_{\mathbf{reactants}}}}$ $\overline{){{\mathbf{K}}}_{{\mathbf{c}}}{\mathbf{=}}\frac{\mathbf{\left[}\mathbf{products}\mathbf{\right]}}{\mathbf{\left[}\mathbf{reactants}\mathbf{\right]}}}$

*Note that solid and liquid compounds are ignored in the equilibrium expression.*

At 700 K, the reaction below has a K_{p} value of 54. An equilibrium mixture at this

temperature was found to contain 1.84 atm of I_{2} and 4.24 atm of HI. Calculate the equilibrium pressure of H_{2}.

H_{2}(g) + I_{2}(g) <=> 2 HI(g). Enter answer to 2 decimal places

A. 0.18 atm

B. 0.04 atm

C. 5.53 atm

D. 0.30 atm

E. 0.81 atm

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