Question

we put 10 moles of n2o into a 2 DM cube container at some temperature where it decomposes according to the reaction that to n2o the reversible reaction to N2 + O2 at equilibrium 2.20 moles of n2o remains calculate the value of KC for the reaction​

Answer 1

Answer:

The equilibrium constant in terms of molar concentration is called the Kc for the reaction. It can be obtained by taking the ratio of the concentrations of the products and reactants for a given system in chemical equilibrium.

Answer 2

Answer:

The value of equilibrium constant is

$K_{C} = 24.51$

Explanation:

Given that,

Initial no.of moles of N2O=10mol

Volume of container is

$V = 2dm {}^{3} = 2lit$

Final no.of moles of N20 at equilibrium=2.20mol

The balanced chemical reaction is

$2N _{2}O = = > 2N_{2} + O_{2} \\ 10mol \: \: \: 0mol \: 0mol \\ (10 - 2x)mol \: \: (2x)mol \: (2x)mol$

We know that at equilibrium

$10 - 2x = 2.20 = > x = 3.90mol$

Now we need to find concentrations of all reactants and products[NO]M

$[ N_{2}O ] = \frac{no \: of \: moles}{volume} = \frac{2.20}{2} = 1.10M$

$[N_{2}] = \frac{2 \times 3.90}{2} = 3.90M \\ [O _{2} ] = \frac{3.90}{2} = 1.95M$

Now we find the value of equilibrium constant by the following relation

$K _{C} = \frac{[N_{2}] {}^{2}[O _{2} ] }{[N _{2} O] {}^{2} }$

By substituting the values we get,

$K _{C} = \frac{3.90 {}^{2} \times 1.95 } {1.10 {}^{2} } = 24.51$

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