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.

Step-by-step 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) * (RT)^(\Delta n)`

`5.6 * 10^(-3) = K_(c) * (0.0821 * 700)^(1)`

`K_(c) = 0.097 * 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) * P_{O_(2)}}{P^(2)_(HgO)}`

`5.6 * 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.