Question

Calculate the partial pressure (in atm) of CH₃OH at equilibrium when 4.84 atm of CH₃I and 4.84 atm of H₂O react at 3000 K according to the following chemical equation:

CH3I (g) + H2O (g) ⇌ CH3OH (g) + HI (g)

Kp = 7.47×101

Report your answer to three significant figures in scientific notation.

**** i have just one chance !!

Answer 1

Kp for the given reaction= [PCH3OH] / [PCH3I] [PH2O]= 7.47*10¹= 74.7

given initally partial pressures of both CH3I and H2O are at 4.84 atm

let x = drop in partial presssrue and x= partial pressure of CH3OH at equilibrium

x/(4.84-x)²= 74.7

This problem can be solved by trial and error by assuming some value of x and matching LHS and RHS

lex= 4.5

LHS= 4.5/(4.84-4.5)² = 38.92 and not the same as RHS

assume some higher value let us sat x=4.6 atm LHS= 79,865 > RHS

is has to be reduced to make it equal to RHS.

using solver of excel and matching LHS and RHS, we get x= 4.592

At equilibrium, the partial pressure of CH3OH= 4.592 atm