Calculate the equilibrium constant Kp for this reaction, given the following information (at 296 K ): 2NO(g)+Br2(g)⇌2NOBr(g)Kc=1.8 2NO(g)⇌N2(g)+O2(g)Kc=2.3×1030

Explanation:

The relation between Kp and Kc is given below:

Where,  

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

Given: Kc = 1.8

Temperature = 296 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(2+1) = -1  

Thus, Kp is:

For the second equilibrium reaction:

Given: Kc =

Temperature = 296 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(2) = 0

Thus, Kp is:

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.