Equilibrium question

[belochka]

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The original r-n is
4A(s)+2B<------->4C(g), k=50
And if:
8A(s)+ 4B(g)<----->8C(g) So, why when it is double of the original, k(new)= 50^2, not 2*50?

Thanks,

 

[ktran17]

It is k^2 rather than x(k), where x= the multiplying factor because of the way k is derived in contrast to something like standard enthalpy. k is derived by [products] at eq/ reactants at eq. Standard enthalpy on the other hand, is the sum of products - sum of reactants. Therefore, when dealing with multiplying factors and equilibrium expressions always yield k^(x).

Another important thing to note, if you want to combine k's, you have to multiply them not add them. Again, this is due to the mathematical derivation of k.

For example, if you have k1 and k2, then k3= k1 x k2. NOT k1+k2.

 

[Dotoday]

This js the formula. When you double the reactants, you square tje equilibrium. When half the reactants, you take the square root if the equilibrium.

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[belochka]

Thanks, guys!