Question

Consider the following chemical equilibrium: 2PbO+O2=2PbO2Now write an equation below that shows how to calculate from for this reaction at an absolute temperature . You can assume is comfortably above room temperature. If you include any common physical constants in your equation be sure you use their standard symbols, found in the ALEKS Calculator.

Answer 1

The question is incomplete, here is the complete question:

Consider the following chemical equilibrium: 2PbO + O₂= 2PbO₂

Now write an equation below that shows how to calculate $K_p$ from $K_c$ for this reaction at an absolute temperature T.

You can assume is comfortably above room temperature. If you include any common physical constants in your equation be sure you use their standard symbols, found in the ALEKS Calculator.

Answer: The expression of $K_p$ from $K_c$ for given equation is written below.

Explanation:

For the given chemical equation:

$2PbO(s)+O_2(g)\rightleftharpoons 2PbO_2(s)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{1}{[O_2]}$

The concentration of pure liquids and solids are taken as 1 in equilibrium constant expression. So, the concentration of lead (II) oxide and lead (IV) oxide do not appear in the expression.

Relation of $K_p\text{ with }K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta n_g}$

Where,

$K_p$ = equilibrium constant in terms of partial pressure

$K_c$ = equilibrium constant in terms of concentration

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

T = absolute temperature = T

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=0-1=-1$

Putting values in above equation, we get:

$K_p=K_c\times (RT)^{-1}\\\\K_p=\frac{K_c}{RT}$

Hence, the expression of $K_p$ from $K_c$ for given equation is written above.