Answer to Question #128909 in General Chemistry for M

Question

Hi,I am unable to submit order.Please help,please may I have help?

The equilibrium constant, Kc is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, Kp calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation
Kp=Kc(RT)^Δn
where R=8.314 L⋅kPa/(K⋅mol) T is the absolute temperature, and Δn is the change in the number of moles of gas (sum moles products - sum moles reactants). For example, consider the reaction.
For the reaction
3A(g)+2B(g)⇌C(g)
Kc= 72.2 at a temperature of 237 ∘C .Calculate the value of Kp
Thank you very much.

Answer 1

$K_p(equilibrium\ constant)\ is \ the\ relationship\ between\ the \ concentrations\ of\newline the\ products\ and \ the\ reactants \ at\ equilibrium\ in \ relaton\ to\ partial\ pressures.$

$K_p=K_c(RT)^{\Delta n}\ \ where,\newline R=8.314L-kPa/(K-mol)\ or0.08514L/K.mol\newline T=237^0\ \ or\ 510K\newline \Delta n=mol\ product\ -\ mol\ of\ reactants.\ 1-5=-4$

$K_p=72.2\ *(0.08314*510)^{-4}$

$K_p=72.2\ *\ (3.1*10^{-7})\newline K_p=2.2*10^{-5}$