Question

At a particular temperature, Kp-483 × 102 for the reaction H2 (g) + 12(g) ? 2H1(g) If 3 01 atm of H-(g) and 3.01 atm of I2(g) are introduced into a 1.00-L- container, calculate the equilibrium partial pressures of all Partial pressure of H2 Partial pressure of I2atm Partial pressure ofHIatm atm

Answer 1

Answer:

P[H2] = P[I2] = 0.02 atm

P[HI] = 5.98 atm

Explanation:

The ICE table for the given reaction is:

                H2(g)   +      I2(g)      ↔     2HI(g)

Initial       3.01atm       3.01atm            -

Change    -x                 -x                    +2x

Equilib     (3.01-x)        (3.01-x)              2x

The equilibrium constant, Kp is given as:

$Kp = \frac{P_{HI}^{2}}{P_{H2}*P_{I2}}$

$483*10^{2} = \frac{(x)^{2}}{(3.01-x)^{2}}$

x = 2.99 atm

Equilibrium partial pressure of H2 = I2 = 3.01-2.99 = 0.02 atm

Equilibrium partial pressure of HI = 2x = 2(2.99) = 5.98 atm

Answer 2

Answer:

The partials pressures are:

PHI = 5.016 atm

PH2 = PI2 = 0.502 atm

Explanation:

please look at the solution in the attached Word file

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