At 920 k, kp = 0.40 for the following reaction. 2 so2(g) + o2(g) equilibrium reaction arrow 2 so3(g) calculate the equilibrium partial pressures of so2, o2, and so3 produced from an initial mixture in which the partial pressures of so2 and o2 = 0.52 atm and the partial pressure of so3 = 0 (exactly).

The equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.

Explanation:

We are given:

Initial partial pressure of sulfur dioxide = 0.52 atm

Initial partial pressure of oxygen = 0.52 atm

Initial partial pressure of sulfur trioxide = 0 atm

For the given chemical reaction:

Initial:                 0.52      0.52

At eqllm:         0.52-2x    0.52-x        2x

The expression of for above equation follows:

We are given:

Putting values in above equation, we get:

Neglecting the negative values because partial pressure cannot be negative.

So, x = 0.077

Equilibrium partial pressure of sulfur dioxide =

Equilibrium partial pressure of oxygen =

Equilibrium partial pressure of sulfur trioxide =

Hence, the equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa

the formula for the conversion of temperature from degree celsius to kelvin is:

the sample of water at is converted to kelvin temperature by using the above formula as:

on comparing the two samples of water at or we can say that at and at respectively. the difference between the temperatures of two samples of water is:

hence, one can say that the sample of water at is greater than the sample of water at by .