Assume that the change in pressure of H2S is small enough to be neglected in the following problem. H2S(g) ⇌ 2H2(g) + S2(g) Kp =2.2 x 10^-6

a. Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H2S with an initial pressure of 0.824 atm.
b. Show that the change is small enough to be neglected.

pH2S = 0.816 atm

pH2 =  0.0153  atm

pS2 = 0.00765 atm

Explanation:

Step 1: Data given

Kp = 2.2 *10 ^-6

Initial pressure H2S = 0.824 atm

Step 2: The equation

H2S(g) ⇌ 2H2(g) + S2(g)

Step 3: The initial pressure

pH2S = 0.824 atm

pH2 = 0 atm

pS2 = 0 atm

Step 4: The pressure at the equilibrium

pH2S = 0.824 - x atm

pH2 = 2x atm

pS2 = x atm

Kp = ((pH2)² * (pS2))/pH2S

2.2 * 10^-6  = ((2x)² * x) / (0.824 - x)

2.2*10^-6 = 4x³ / (0.824 - x)

4x³ = 2.2*10^-6 * (0.824 - x)

x = 0.00765

pH2S = 0.824 - 0.00765 = 0.816 atm

pH2 = 2* 0.00765 = 0.0153  atm

pS2 = 0.00765 atm

The change is small enough to be neglected because 0.00765 << 0.824

If we calculate the result with or without this change the result will not be very different

2.2*10^-6 = 4x³ / (0.824 )

x = 0.00768 atm

pH2S = 0.824 - 0.00768 = 0.816 atm

pH2 = 2* 0.00768 = 0.0154  atm

pS2 = 0.00768 atm

f=ma newton

f=100x2.5=250n