[Landy813]

Hello all i need some help with this,

The oxidation of benzhydrol, (C6H5)2CHOH, in a basic solution of potassium permanganate, KMnO4, has the following time-independent stoichometry:

(C6H5)2CHOH(aq) + 2MnO4 (aq) + 2OH−(aq) = (C6H5)2CO(aq) + 2MnO4 2−(aq) + 2H2O  (reaction 1)

The reaction was followed by monitoring the decrease in benzhydrol concentration (we will refer to this as ‘BH’ for short) as a function of time. The results of an experiment carried out at 298.2 K are given in Table 1. The initial concentrations of reactants are as follows:

[BH]₀ = 3.9 × 10−4 mol dm−3
[MnO₄ ]₀ = 7.8 × 10−4 mol dm−3
[OH⁻]₀ = 2.2 × 10−2 mol dm−3

For Reaction 1 the initial concentration of OH− was 2.2 × 10−2 mol dm−3and the partial orders with respect to BH and MnO₄ are equal to 1. Thus, if the concentrations of BH and MnO₄ ⁻ are related as follows:

[BH]₀ = [MnO₄]₀/2

show how the general rate equation you suggested in part (c) can be
simplified to give the following pseudo-order rate equation


Can anyone tell me what the number 2 means in this equation

J= 2K_R'[BH]²

is it there are 2 reactants in excess if i use the the initial rate method.

Regards

Landy

[Hunt]

The "2" is added because in this case [MnO₄^-] = 2[BH] at any time.


Because the concentration of hydroxide ions is much greater than the concentrations of other reactants , you can do an approximation and assume [OH]^- is a constant. Then the rate J = k'[BH][MnO₄^-]

If [BH] = [BH]_o - x then

[MnO₄^-] = [MnO₄^-] _o - 2x

Initially 2 [BH]_o = [MnO₄^-]_o

[MnO₄^-] = 2 [BH]_o - 2x = 2 [BH]

so J = 2k' [BH]²

[renge ishyo]

I think it refers to a substitution made (the two comes from the stoichiometry of the balanced chemical equation).

If the initial equation can be written as (assuming first order for both BH and Mn04):

J = K_R[BH]₀[Mn0₄]₀

using the equation based on the stoichiometry to solve for the concentration of Mn0₄:

[Mn0₄]₀ = 2[BH]₀

substitute in for [Mn0₄]₀ in J:

J = K_R[BH]₀(2[BH]₀) or upon rearrangement:

J = 2K_R[BH]₀²

Edit: Blarg, too slow I guess