Question

The vapour pressure of mercury is 0.002 mm Hg at 27°C . Kc for the process : Hg(l) --> Hg(g) is ?

Answer 1

Answer:1.068×10^-7

Explanation:Hg(l)=Hg(g)

Kp=PHg=0.002mm Hg=0.002/760 atm

Kp =KC(RT)^∆ng

KC=Kp/RT=1.068×10^-7

Answer 2

The value of $K_c$ for the given equation is $1.07\times 10^{-7}$

Explanation:

For the given chemical equation:

$Hg(l)\rightleftharpoons Hg(g)$

The expression of $K_p$ for above equation follows:

$K_p=p_{Hg(g)}$

The partial pressure of solids and liquids are taken as 1 in the equilibrium constant expression.

We are given:

$p_{Hg(g)}=0.002mmHg$

So, $K_p=0.002$

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 = 0.002

$K_c$ = equilibrium constant in terms of concentration

R = Gas constant = $62.364\text{ L. mmHg }mol^{-1}K^{-1}$

T = temperature = $27^oC=27+273=300K$

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

Putting values in above equation, we get:

$0.002=K_c\times (62.364\times 300)^{1}\\\\K_c=1.07\times 10^{-7}$

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