Evaluate the expression6^2 + x- x ^2 for x =3
1 answer:
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For the reaction,
Initial concentration:
<u>0.1M</u> , <u>0.1M</u> , <u>0</u> , <u>0</u>
Let 'x' mole per litre of each of theproduct be formed.
At equilibrium:
<u>(0.1 – x)M</u> , <u>(0.1 – x)M</u> , <u>xM</u> , <u>xM</u>
where x is the amount of Carbon dioxide and Hydrogen, at equilibrium.
Hence, equilibrium constant can be written as,
→ x² = 4.24 (0.01 + x² – 0.2x)
→ x² = 0.0424 + 4.24 x² - 0.848x
→ 3.24x² - 0.848x + 0.0424 = 0
<em>a = 3.24, b = -0.848, c = 0.0424</em>
(for quadratic equation ax² + bx+c=0)
Here, the value 0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration.
∴ The equilibrium concentrations are :-