Chapter 3: Rate Laws
Rate Law for Reversible Reactions
Write the rate law for the elementary reaction
\(\text{A} \overset{k_{fA}}{\underset{k_{rA}}{\rightleftharpoons}} 2\text{B}\)
Here kfA and krA are the forward and reverse specific reaction rates both defined with respect to A.
Hint: What is the net rate of reaction?
$ (a) r_{A} = k_{fA}C_{A} - k_{rA}{C_{B}}^2 $
$ (b) r_{A} = k_{rA}{C_{B}}^2 - k_{f}{C_{A}} $
$ (c) r_{A} = k_{fA}(C_{A} - \frac{{C_{B}}^2}{k_{c}}) $
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What is the net rate of reaction?
(1) \(\text{A} \xrightarrow{k_{fA}} 2\text{B}\) \(-r_{Af} = k_{fA} C_A\)
(2) \(2\text{B} \xrightarrow{k_{rA}} \text{A}\) \(r_{Ar} = k_{rA} C_B^2\)
\(\text{rate}_{\text{net}} = \text{rate}_{\text{forward}} + \text{rate}_{\text{reverse}}\)
Solution
Write the rate law for the elementary reaction
\(\text{A} \overset{k_{fA}}{\underset{k_{rA}}{\rightleftharpoons}} 2\text{B}\)
Here kfA and krA are the forward and reverse specific reaction rates both defined with respect to A.
What is the net rate of reaction?
(1) \(\text{A} \xrightarrow{k_{fA}} 2\text{B}\) \(-r_{Af} = k_{fA} C_A\)
(2) \(2\text{B} \xrightarrow{k_{rA}} \text{A}\) \(r_{Ar} = k_{rA} C_B^2\)
\(\text{rate}_{\text{net}} = \text{rate}_{\text{forward}} + \text{rate}_{\text{reverse}}\)
\(r_A = -k_f C_A + k_r C_B^2\)
\(-r_A = k_{fA} C_A - k_r C_B^2\)
\(= k_{fA} \left[C_A - \frac{C_B^2}{k_f / k_r}\right]\)
\[ \boxed{-r_A = k_{fA} \left[C_A - \frac{C_B^2}{K_c}\right]} \]
At equilibrium
\(-r_A \equiv 0\)
\[ \boxed{K_c = \frac{C_{Be}^2}{C_{Ae}}} \]