Question

For the given elementary reactions, the steady-state concentration of X is

• A. \(\frac{K_1[P]^2+K_3[Q][R]}{K_2+K_4}\)
• B. \(\frac{\frac{1}{2}K_1[P]^2+k_3[Q][R]}{k_2+k_4}\)
• C. \(\frac{k_1[p]+k_3[Q][R]}{k_2+k_4}\)
• D. \(\frac{K_1[P]+k_3[Q][R]}{k_2+k_4}\)

Answer 1

The Correct Option is A

Step 1: Write the Rate Laws 

For the reaction \( 2P \xrightarrow{k_1} X \xrightarrow{k_2} Q + R \), the rate laws for the elementary reactions are:

• The formation of \( X \) from \( P \): \[ \text{Rate of formation of } X = k_1 [P]^2 \]
• The consumption of \( X \) to form \( Q \) and \( R \): \[ \text{Rate of consumption of } X = k_2 [X] \]

Step 2: Applying the Steady-State Assumption

At steady-state, the rate of formation of \( X \) is equal to the rate of consumption of \( X \). Therefore, we have:

\[ k_1 [P]^2 = k_2 [X] \] This equation can be rearranged to solve for the concentration of \( X \) in terms of \( [P] \): \[ [X] = \frac{k_1 [P]^2}{k_2} \]

Step 3: Accounting for Additional Reactions

If additional reactions contribute to the production or consumption of \( X \), such as the reaction \( X \xrightarrow{k_3} Q + R \), we can include these as follows:

\[ k_1 [P]^2 + k_3 [Q][R] = k_2 [X] + k_4 [X] \]

Step 4: Solving for the Steady-State Concentration of \( X \)

Now, solving for the steady-state concentration of \( X \) considering both reactions: \[ [X] = \frac{k_1 [P]^2 + k_3 [Q][R]}{k_2 + k_4} \]

Final Answer:

The steady-state concentration of \( X \) is given by:

\(\frac{K_1 [P]^2 + K_3 [Q][R]}{K_2 + K_4}\)