In our chemistry textbook, the equilibrium constant $K_c$ is derived this way:

$$\ce{aA + bB <=> cC + dD}$$

in this reaction, rate of forward reaction is $k_f= k_1[A]^a[B]^b$

rate of backward reaction is $k_b = k_2[C]^c[D]^d$

in equilibrium, equalizing the two rates we get $$ K_c = \frac{k_1}{k_2}=\frac{[A]^a[B]^b}{[C]^c[D]^d}$$

Now, there is a problem... we know the $K_c$ depends on how we equalize the reaction... like we can write $$\ce{H2 +.5O2 -> H2O}$$ or $$\ce{2H2 + O2 -> 2H2O}$$, etc ways. But how the rate of reaction depends on the concentration? to what power?

In our chemistry textbook, the equilibrium constant $K_c$ is derived this way:

$$\ce{aA + bB <=> cC + dD}$$

in this reaction, rate of forward reaction is $k_\mathrm f= k_1[A]^a[B]^b$

rate of backward reaction is $k_\mathrm b = k_2[C]^c[D]^d$

in equilibrium, equalizing the two rates we get $$ K_c = \frac{k_1}{k_2}=\frac{[A]^a[B]^b}{[C]^c[D]^d}$$

Now, there is a problem... we know the $K_c$ depends on how we equalize the reaction... like we can write $$\ce{H2 + 0.5O2 -> H2O}$$ or $$\ce{2H2 + O2 -> 2H2O}$$, etc ways. But how the rate of reaction depends on the concentration? to what power?

In our chemistry textbook, the equilibrium constant Kc is derived this way:

aA+bB=cC+dD

in this reaction, rate of forward reaction is= k1*[A]^a*[B]^b

rate of backward reaction= k2*[C]^c*[D]^d

in equilibrium, equalizing the two rates we get k1/k2=Kc=[A]^a*[B]^b/[C]^c*[D]^d

Now, there is a problem.. we know the Kc depends on how we equalize the reaction.. like we can write H2+.5O2=H2O or 2H2+O2=2H20 etc ways. But how the rate of reaction depends on the concentration? to what power?

In our chemistry textbook, the equilibrium constant $K_c$ is derived this way:

$$\ce{aA + bB <=> cC + dD}$$

in this reaction, rate of forward reaction is $k_f= k_1[A]^a[B]^b$

rate of backward reaction is $k_b = k_2[C]^c[D]^d$

in equilibrium, equalizing the two rates we get $$ K_c = \frac{k_1}{k_2}=\frac{[A]^a[B]^b}{[C]^c[D]^d}$$

Now, there is a problem... we know the $K_c$ depends on how we equalize the reaction... like we can write $$\ce{H2 +.5O2 -> H2O}$$ or $$\ce{2H2 + O2 -> 2H2O}$$, etc ways. But how the rate of reaction depends on the concentration? to what power?