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What does a KC less than 1 mean?

When Kc is less than 1, reactants exceed products. When much less than 1 (Kc can never be negative...so when it is close to zero) the reaction hardly occurs at all.

en.wikibooks.org - IB Chemistry/Equilibrium - Wikibooks, open books for an open world
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Equilibrium

Equilibrium primarily deals with reversible reactions and the state of equilibrium itself. Equilibrium can be defined as a state reached of a reversible reaction, in which the rate of the forward and backward reactions equal.

Dynamic equilibrium [ edit | edit source ]

8.1.1 : In all reactions, there are in fact two reactions occurring, one where the reactants produce the products, and the other where the products react to form the reactants. In some reactions, this second reaction is insignificant, but in others there comes a point where the two reactions exactly cancel each other out...thus the reactants and products remain in equal proportions, though both are continually being used up and produced at the same time.

8.2 The position of equilibrium

8.2.1 : The equilibrium constant Kc is a constant which represents how far the reaction will proceed at a given temperature. 8.2.2 : When Kc is greater than 1, products exceed reactants (at equilibrium). When much greater than 1, the reaction goes almost to completion. When Kc is less than 1, reactants exceed products. When much less than 1 (Kc can never be negative...so when it is close to zero) the reaction hardly occurs at all. 8.2.3 : The only thing which can change the value of Kc for a given reaction is a change in temperature. The position of equilibrium, however, can change without a change in the value of Kc. Effect of Temperature : The effect of a change of temperature on a reaction will depend on whether the reaction is exothermic or endothermic. When the temperature increases, Le Chatelier's principle says the reaction will proceed in such a way as to counteract this change, ie lower the temperature. Therefore, endothermic reactions will move forward, and exothermic reactions will move backwards (thus becoming endothermic). The reverse is true for a lowering of temperature. Effect of Concentration : When the concentration of a product is increased, the reaction proceeds in reverse to decrease the concentration of the products. When the concentration of a reactant is increased, the reaction proceeds forward to decrease the concentration of reactants. Effect of Pressure : In reactions where gases are produced (or there are more mols of gas on the right), an increase in pressure will force the reaction to move to the left (in reverse). If pressure is decreased, the reaction will proceed forward to increase pressure. If there are more mols of gas on the left of the equation, this is all reversed. 8.2.4 : Based on the previous section, you should be able to predict what's going to happen given a reaction if the temperature, pressure, or concentration is changed. 8.2.5 : A catalyst does not affect either Kc or the position of equilibrium, only the rate of reaction. 8.2.6 : N2(g) + 3H2(g) <=> 2NH3(g) : delta-H = -92.4 kJ mol-1 as can be seen, there are more mols of gas on the left than the right, so a greater yield will be produced at high pressure. The reaction is exothermic, therefore it will give a greater yield at low temperatures, however this is not possible as the rate of reaction becomes too low, and the temperature must actually be increased. A catalyst of finely divided iron is also used to help speed the reaction (finely divided to maximize the surface area).

HL Material [ edit | edit source ]

Topic 17 is the additional HL material for Topic 8.

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What happens when the value of K is negative?

For negative k, you'll get a spiral that spirals out anti-clockwise.

math.stackexchange.com - exponential function - $y=k^x$ when $k$ is negative - Math Stack Exchange

$egingroup$

Lets consider $k^x$ when x is an integer and $k<1$.

Provided $x$ is an integer $x in mathbb{I}$ we have no issues.

Taking for example $k = -2$ we have:

$$ egin{array}{c|c|c} x& ext{}&k^x hline 3&(-2)^3&-8 2&(-2)^2&4 1&(-2)^1&-2 0&(-2)^0&1 -1&dfrac{1}{(-2)^1}&-dfrac{1}{2} -2&dfrac{1}{(-2)^2}&dfrac{1}{4} -3&dfrac{1}{(-2)^3}&-dfrac{1}{8} end{array} $$

Now consider $x = dfrac{1}{2}$, and $k^x = sqrt{-2} = ?$

We can solve this with complex numbers $k^x = i cdot sqrt{2}$

Where $i = sqrt{-1}$

Similarly $x = -dfrac{1}{2}$, and $k^x = dfrac{1}{sqrt{-2}} = - dfrac{i}{sqrt{2}}$

But our solutions are not real numbers any more $k^x

otin mathbb{R}$

When $ k lt 0$ we do not have real solutions for all x

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