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

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:

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

the maximum mass of nh3 is nh3 in grams