At 700 k, kp for the following equilibrium is (5.6 x 10-3) 2hgo(> 2hg(l) + o2(g) suppose 51.2 g of mercury(ii) oxide is placed in a sealed 3.00-l vessel at 700 k. what is the partial
pressure of oxygen gas at equilibrium? (r = 0.0821 lxatm/(kxmol))
a) 0.075 atm
b) 0.0056 atm
c) 4.5 atm
d) 19 atm
e) 2.3 atm

Option (B) is the correct answer.

Explanation:

According to the given reaction equation, formula to calculate is as follows.

    = coefficients of gaseous products - gaseous reactants                    

                  = 1 - 0

                  = 1

Also we know that,

For the equation,

Activity of solid and liquid = 1

As,    

Hence, = 0.0056 atm

Thus, we can conclude that partial pressure of oxygen gas at equilibrium is 0.0056 atm.

0

explanation:

488479

rvmoo+478)9.68