Q.The rate constants of the forward reaction and reverse reaction are 0.05 $sec^{-1}$ and 2 $sec^{-1}$ respectively, then $K_c$ is :

The general unit of a rate constant is $(\text{mol L}^{-1})^{1-n}\,s^{-1}$ for a reaction of order $n$; this reduces to plain $s^{-1}$ (no concentration term at all) only when $n=1$, i.e. for a first-order reaction.

For a reaction of order $n$, rate $=k[\text{conc}]^n$; since rate always has units of $\text{mol L}^{-1}s^{-1}$, rearranging gives $k$'s units as $(\text{mol L}^{-1})^{1-n}s^{-1}$. Checking each order: zero order ($n=0$) gives $k$ units of $\text{mol L}^{-1}s^{-1}$; first order ($n=1$) gives $(\text{mol L}^{-1})^0s^{-1}=s^{-1}$ (concentration term vanishes entirely); second order ($n=2$) gives $\text{mol}^{-1}Ls^{-1}$; third order ($n=3$) gives $\text{mol}^{-2}L^2s^{-1}$. Since the given rat …

At equilibrium the forward and reverse rates are equal, so the equilibrium constant equals the ratio of the forward to the reverse rate constant: $K_c = k_f/k_r$.

For a reversible reaction $A \rightleftharpoons B$, at equilibrium the rate of the forward reaction equals the rate of the reverse reaction:

$k_f[A] = k_r[B]$

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