We’re being asked to calculate the equilibrium pressure of H_{2}, given the following reaction:

**H _{2}(g) + I_{2}(g) ⇌ 2 HI(g) K_{p} = 54**

We're given the following equilibrium pressures::

P_{I}_{2} = 1.49 atm

P_{HI} = 4.08 atm

The **K _{p} expression** for the reaction is:

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

temperature was found to contain 1.49 atm of I_{2} and 4.08 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. 0.05 atm

b. 0. 11 atm

c. 0.21 atm

d. 2.07 atm

e. 603.29 atm

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