[Gerard]

we are asked to derived the rate equation of
A+2B=R
yes i derived the other half which is
-ln{(M-2XA)/M(1-XA)}=CA0(M-2)KT
but then we are asked what if M=2 this is where our problem arises..
what if M=2
what will happen to the equation???further more we are also asked what are the thermodynaimc quantities that can predict whether a reaction is reversible or irreversible???

[Donaldson Tan]

A + 2B => R

I assume XA refers to X_A - the conversion of A.

This is the rate equation: Rate = - d[A]/dt = K[A][ B ]²

- d[A]/dt = - (d/dt)([A]₀(1 - X_A) = [A]₀dX_A/dt

From the stoichiometric ratio, [ B ]₀X_B = 2 [A]₀X_A

[ B ] = [ B ]₀(1 - X_B)
 
K[ A ][ B ]² = K[A]₀(1 - X_A)([ B ]₀-[ B ]₀X_B)² = K[A]₀(1 - X_A)([ B ]₀ - 2[A]₀X_A)²

[A]₀dX_A/dt = K[A]₀(1 - X_A)([ B ]₀ - 2[A]₀X_A)²

dX_A/dt = K(1 - X_A)([ B ]₀ - 2[A]₀X_A)²

Solving the above ODE will yield the correct reaction profile.

[Donaldson Tan]

further more we are also asked what are the thermodynaimc quantities that can predict whether a reaction is reversible or irreversible???

dG = dH - T.dS

if dG = 0, then your reaction is reversible

[Gerard]

further more we are also asked what are the thermodynaimc quantities that can predict whether a reaction is reversible or irreversible???

dG = dH - T.dS

if dG = 0, then your reaction is reversible

i feel ashamed of myself! i did not took acount the gibbs free energy what ever happened to my chemical thermodynamics
thank you
a lot!