Given the equation :
CH3OH +Cl2 →← 2 CH3CL + 2OH ^-1
at equilibrium concentration, we have:
(1.5 -2x )/ 5L + (1.0 -x )/5L =2x /5L + 2x /5L
→ (1.5-2x+ 1 -x )/ 5L = 4x /5L
→(0.5 -3x)/5L = 4x/5L (5L in the Left hand side cancels 5L inthe right hand side)
→ 0.5 = 4x +3x
0.5 = 7x
x = 0.5/7
x = 0.071
( Extra note : Now lets test and see if this balances ):
LHS : (1.5 -2x )/ 5L + (1.0 -x )/5L = 0.5 -3x)/5L
= 0.5 - 3(0.071)/5
= 0.287 /5 = 0.05
RHS :4(x) /5 = 4(0.071) /5
= 0.28/5 = 0.05
at equlibrium , LHS = RHS therefore our x value is 0.07
