At a certain temperature, the K p Kp for the decomposition of H 2 S H2S is 0.834 . 0.834. H 2 S ( g ) − ⇀ ↽ − H 2 ( g ) + S ( g ) H2S(g)↽−−⇀H2(g)+S(g) Initially, only H 2 S H2S is present at a pressure of 0.150 atm 0.150 atm in a closed container. What is the total pressure in the container at equilibrium?

Total pressure = 0.2798 atm

Explanation:

Step 1: Data given

Kp = 0.834

Initially, only H2S is present at a pressure of 0.150 atm

Step 2: The balanced equation

H2S(g)⇄ H2(g) + S(g)

Step 3: The initial pressure

pH2S = 0.150 atm

pH2 = 0 atm

pS = 0 atm

Step 3: The pressure at the equilibrium

For 1 mol H2S we'll have 1 mo lH2 and 1 mol S

pH2S = 0.150 - X atm

pH2 = X atm

pS = X atm

Step 4: Calculate Kp

Kp = (pS*pH2)/(pH2S)

0.834 = X² / (0.150 - X)

X = 0.1298

pH2S = 0.150 - 0.1298 = 0.0202 atm

pH2 = 0.1298 atm

pS =0.1298 atm

Kp = (0.1298*0.1298) / 0.0202

Kp = 0.834

The total pressure = pH2S + pH2 + pS

Total pressure = 0.0202 atm+ 0.1298 atm+ 0.1298 atm

Total pressure = 0.2798 atm

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6