Question

An engineer examining the oxidation of SO 2 in the manufacture of sulfuric acid determines that Kc = 1.7 x 108 at 600 K:

Answer 1

Answer: $p_{SO_2}=0.017atm$

Explanation:

We are given:

$K_c=1.7\times 10^8$

Relation of $K_p$ with $K_c$ is given by the formula:

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

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 = temperature = $600K$

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=1.7\times 10^8\times (0.0821\times 600)^{-1}\\\\K_p=3.4\times 10^6$

The chemical reaction follows the equation:

                 $2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression for $K_p$ for the given reaction follows:

$K_p=\frac{(p_{SO_3})^2}{ p_{O_2}\times {(p_{SO_2})^2}}$

We are given:

$p_{SO_3}=300atm$   $p_{O_2}=100atm$

Putting values in above equation, we get:

$3.4\times 10^6=\frac{(300)^2}{100\times {(p_{SO_2})^2}}$

$p_{SO_2}=0.017atm$

Hence, the partial pressure of the $SO_2$ at equilibrium is 0.017 atm.