At a certain temperature, 0.4011 mol of n2 and 1.621 mol of h2 are placed in a 3.50 l container. n2(g)+3h2(g)↽−−⇀2nh3(g) at equilibrium, 0.1001 mol of n2 is present. calculate the equilibrium constant, kc .

The equilibrium constant ≈120.07

Explanation:

Step 1: The balanced equation

N2(g)+3H2(g)↽−−⇀2NH3(g)

This means that for 1 mole of N2 consumed, there is needed 3 moles of H2 to produce 2 moles of NH3

Step 2: Calculate initial moles

Reactants:

 ⇒N2: 0.4011 mol es

 ⇒H2: 1.621 mol es

Products

 ⇒NH3: 0 mol  

Step 3: Calculate number of moles that changed

N2: 0.4011 mol - 0.1001 mol = 0.3010 mol

H2: 3 * 0.3010 mol = 0.9030 mol

NH3: 2 * 0.3010 mol = 0.6020 mol

Step 4: Calculate moles at the equilibrium

N2: 0.1001 moles

 ⇒[N2] = 0.1001moles / 3.5L = 0.0286 M

H2: 1.621 moles -0.9030 moles  = 0.718 moles

 ⇒[H2] = 0.718 moles/ 3.5L  : 0.205 M

NH3: 0.6020 mol es

 ⇒[NH3] =  0.6020 / 3.5= 0.172M

Step 5: Calculate Kc

Kc = [NH3]² / ([N2] [H2]³)

Kc= 0.172²/(0.0286*0.205³)

Kc = 0.029584‬/(0.0286* 0.008615125‬)

Kc  ≈120.07

The Kcis a high value, this shows the position of the equilibrium is to the right.

The equilibrium constant ≈120.07

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.

you gotta add the following molecules so we can give u an answer

explanation: