In which case does the reaction go farthest to completion ?
(i)`K_1=10^15` , (ii)`K_2=10^3` , (iii)`K_3=10^(-5)` ?

Answer

To determine in which case the reaction goes farthest to completion, we need to analyze the equilibrium constants (K) provided for the three scenarios:

1. **Given Values:**
- (i) \( K_1 = 10^{15} \)
- (ii) \( K_2 = 10^{3} \)
- (iii) \( K_3 = 10^{-5} \)

2. **Understanding Equilibrium Constant (K):**
- The equilibrium constant (K) indicates the extent to which a reaction proceeds to completion. A larger K value means that the reaction favors the formation of products, while a smaller K value indicates that the reactants are favored.
- Generally, if \( K > 10 \), the reaction goes significantly towards the products, while if \( K < 10^{-1} \), it favors the reactants.

3. **Comparing the Values:**
- \( K_1 = 10^{15} \): This is a very large value, indicating that the reaction goes almost to completion and favors the products heavily.
- \( K_2 = 10^{3} \): This is also a large value, but not as large as \( K_1 \). It indicates that the reaction favors the products but to a lesser extent than \( K_1 \).
- \( K_3 = 10^{-5} \): This is a very small value, indicating that the reaction favors the reactants significantly and does not proceed far towards completion.

4. **Conclusion:**
- Among the three cases, \( K_1 = 10^{15} \) is the largest. Therefore, the reaction in this case goes farthest to completion.

**Final Answer:** The reaction goes farthest to completion in case (i) where \( K_1 = 10^{15} \).

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