CH3 + HCl <=> CH3Cl + H2O
Kp = 4.7 x 10^3 at 400K.

CH3Cl and HCl combine in 10.00L at 400K. The pressure of CH3OH is 0.250 atm and the pressure of HCl is 0.600 atm (I might have wrote this down wrong).

Does the pressure increase, decrease, or stay the same if equilibrium approaches?

Using Kp, calculate the final partial pressure of HCl at equilibrium.

A student claims the final partial pressure is small but not zero. Justify or argue against this claim and explain why.

The pressure of CH3OH and HCl will decrease.

The final partial pressure of HCl is 0.350038 atm

Explanation:

Step 1: Data given

Kp = 4.7 x 10^3 at 400K

Pressure of CH3OH = 0.250 atm

Pressure of HCl = 0.600 atm

Volume = 10.00 L

Step 2: The balanced equation

CH3OH(g) + HCl(g) <=> CH3Cl(g) + H2O(g)

Step 3: The initial pressure

p(CH3OH) = 0.250atm

p(HCl) = 0.600 atm

p(CH3Cl)= 0 atm

p(H2O) = 0 atm

Step 3: Calculate the pressure at the equilibrium

p(CH3OH) = 0.250 - X atm

p(HCl) = 0.600 - X atm

p(CH3Cl)= X atm

p(H2O) = X atm

Step 4: Calculate Kp

Kp = (pHO * pCH3Cl) / (pCH3* pHCl)

4.7 * 10³ =  X² /(0.250-X)(0.600-X)

X = 0.249962

p(CH3OH) = 0.250 - 0.249962 = 0.000038 atm

p(HCl) = 0.600 - 0.249962 = 0.350038 atm

p(CH3Cl)= 0.249962 atm

p(H2O) = 0.249962 atm

Kp = (0.249962 * 0.249962) / (0.000038 * 0.350038)

Kp = 4.7 *10³

The pressure of CH3OH and HCl will decrease.

The final partial pressure of HCl is 0.350038 atm

the formula would be mgso4

mgso[subscript]4

1 a

2 d

3 a

4 c

explanation: