consider the reaction between iodine gas and chlorine gas to form iodine monochloride:
i2(g)+cl2(g)? 2icl(g)kp=81.9 (at 298 k)
a reaction mixture at 298 k initially contains pi2=0.35atm and pcl2=0.35atm .what is the partial pressure of iodine monochloride when the reaction reaches equilibrium?

also, consider the reaction of a to form b:
2a(g)? b(g)kc=2.1

The partial pressure of iodine monochloride when the reaction reaches equilibrium is 0.5733 atm

Explanation:

Step 1: Data given

Kp = 81.9

PI2 = 0.35 atm

PCl2 = 0.35 atm

Step 2: The balanced equation

I2(g)+Cl2(g)⇌2ICl(g)

Step 3: The initial pressure

I2 = 0.35 atm

PCl2 = 0.35 atm

PICl = 0 atm

Since the mol ratio for I2: Cl2 is 1:1 there will react on both I2 and Cl2 1X

For every mol of I2 we get 2 mol of ICl

This means there will react +2X

Step 4: pressure at the equilibrium

I2 = (0.35 - X) atm

PCl2 = (0.35 -X) atm

PICl = 2X

Step 5: Calculate partial pressure

Kp = 81.9 = (pICl)² / (PI2*PCl2)

81.9 = (2X)² / ((0.35 - X)*(0.35 - X))

4X² = 81.9 * (0.35 - X)²

4X² = 81.9 *(0.1225 - 0.7X + X²)

0 = 77.9X² - 57.33X + 10.03275

X = 0.28665

I2 = (0.35 - X) = 0.06335 atm

PCl2 = (0.35 -X) = 0.06335 atm

PICl = 0.5733 atm

The partial pressure of iodine monochloride when the reaction reaches equilibrium is 0.5733 atm

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.