Question

At a certain temperature a 23 L container hold 4 gases at equilibrium. The masses are

3.5 g SO3

4.6 g SO2

15.2 g N2

0.98g N2O

What is the value of the equilibrium constant at his temperature for the reaction of SO2 with N2O to form SO3 and N2 ? (Kc =?)

Answer 1

we know the molarity can be calculated by using

MM= molar mass

v= volume = 23L=23000 mL(given)

w= given mass

In the given question , the equilibrium reaction is

and , Kc=equilibrium constant

For 3.5 g of SO₃, the molarity can be calculated as:

Here MM=80 gm/mol

V=23000 mL

w=3.5 g

Now apply

For 4.6 g of SO₂, the molarity can be calculated as:

Here MM=64 gm/mol

V=23000 mL

w=4.6 g

Now apply

For 15.2 g of N₂, the molarity can be calculated as:

Here MM=28gm/mol

V=23000 mL

w=15.2 g

Now apply

For 0.98 g of N₂O, the molarity can be calculated as:

Here MM=44 gm/mol

V=23000 mL

w=0.98 g

Now apply

Now apply

Hence the .