Question

The equilibrium constant for the reaction

Answer 1

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

Answer 2

Answer:

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

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.

$Keq = \frac{[NO][SO3]}{[SO2][NO2]}\\Keq *[SO2][NO2] = [NO][SO3]$

$[NO] = Keq \frac{[SO2][NO2]}{[SO3]}$

Replacing the values given in the problem

$[NO] = (10.8) \frac{[0.07M][1.3M]}{[1.7M]} = 0.578 M$

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