Question

At 700 K, Kp for the following equilibrium is (5.6 x 10-3) 2HgO(s)--> 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

Answer 1

Answer: Option (B) is the correct answer.

Explanation:

According to the given reaction equation, formula to calculate $\Delta n$ is as follows.

   $\Delta n$ = coefficients of gaseous products - gaseous reactants                    

                  = 1 - 0

                  = 1

Also we know that,

        $K_{p} = K_{c} \times (RT)^{\Delta n}$

         $5.6 \times 10^{-3} = K_{c} \times (0.0821 \times 700)^{1}$

             $K_{c} = 0.097 \times 10^{-3}$

For the equation, $2HgO(s) \rightarrow 2Hg(l) + O_{2}(g)$

Activity of solid and liquid = 1

As,     $K_{p} = \frac{P^{2}_{Hg} \times P_{O_{2}}}{P^{2}_{HgO}}$

          $5.6 \times 10^{-3} = P_{O_{2}}$

Hence, $P_{O_{2}}$ = 0.0056 atm

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