Question

The equilibrium constant for the reaction

SO2(g)+NI2(g)⇌NO(g) +SO3(g) is Keq= ([NO][SO3])/([SO2][NO2]).

What would the equilibrium concentration of NO be in a gas mixture that contains 1.7 M SO3, 0.070 M SO2, and 1.3 M NO2 if Keq=10.8 at those conditions

Answer 1

The equilibrium concentration of NO in the gas mixture should be 0.578 M

Step-by-step explanation:

We have equilibrium data for the chemical reaction in the gas phase:

SO2(g)+NO2(g)⇌NO(g) +SO3(g)

The equilibrium constant (Keq) for this reaction is 10.8.

Data for the components involved in this equilibrium are:

[SO3] = 1.7 M

[NO] = Unknown

[NO2] = 1.3 M

[SO2] = 0.07 M

We can express the Equilibrium constant as the multiplication of the concentrations of products divided by the multiplication of the concentration of reagents. After that, we can isolate [NO] concentration and find its value.

`Keq = ([NO][SO3])/([SO2][NO2])\\Keq *[SO2][NO2] = [NO][SO3]`

`[NO] = Keq ([SO2][NO2])/([SO3])`

Replacing the values given in the problem

`[NO] = (10.8) ([0.07M][1.3M])/([1.7M])  = 0.578 M`

Finally, the equlibrium concentration is [NO] = 0.578 M

Answer 2

Equilibrium concentration of SO3=1.7M

Equilibrium concentration of SO2=0.070 M

Equilibrium concentration of NO2=1.3 M

Equilibrium constant is given as=10.8

The equilibrium reaction is given as:

SO2(g)+NI2(g)⇌NO(g) +SO3(g)

Keq= ([NO][SO3])/([SO2][NO2])

To calculate concentration of NO rearranging the above equation will give:

NO=Keq*[SO2][NO2]/[SO3]

NO=(10.8*0.070*1.3)/1.7

= 0.578M

So concentration of NO will be 0.578M