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

Tthe value of

Explanation:

Initial moles of hydrogen peroxide  = 1.75 mole

Volume of the container = 25 L

Concentration of a substance is calculated by:

So, concentration of hydrogen peroxide =

The given chemical equation follows:

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 =

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:

The expression of for the above reaction follows:

Putting values in above expression, we get:

Hence, the value of

the answer is option a.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally