Question

Hydrogen peroxide may decompose to form water and oxygen gas according to the following reaction 2H202(g) 2H20(g)+02(g) In a particular experiment, 1.75 moles of Hy02 were placed in a 25-L reaction chamber at 307eC After equilibrium was reached, 1.20 moles of H202 remained What is Ke for the reaction?
A. 2.4×10^-3
B. 2.0×10^-4
C. 5.5×10^-3
D. 2.3×10^-2

Answer 1

Answer:

Tthe value of $K_{eq}=2.31\times 10^{-3}$

Explanation:

Initial moles of hydrogen peroxide  = 1.75 mole

Volume of the container = 25 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of hydrogen peroxide = $\frac{1.75}{25 L}=0.07 M$

The given chemical equation follows:

$2H_2O_2(g)\rightleftharpoons 2H_2O(g)+O_2(g)$

Initially 0.07 M                0   0

At equilibrium

(0.07-2x) M                      2x   x

We are given:

Equilibrium moles of hydrogen peroxide  = 1.20 mole

Equilibrium concentration of hydrogen peroxide = $\frac{1.20}{25 L}=0.048 M$

Evaluating the value of 'x', we get:

(0.07-2x) M = 0.048 M

x = 0.011 M

So, the equilibrium concentrations of water vapors and oxygen gas are:

$[H_2O]=2x=2\times 0.011 M= 0.022 M$

$[O_2]=x=0.011 M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^2\times [O_2]}{ [H_2O_2]^2}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.022 M)^2\times (0.011 M)}{(0.048 M)^2}\\\\K_{eq}=2.31\times 10^{-3}$

Hence, the value of $K_{eq}=2.31\times 10^{-3}$