If the coefficients of a given reaction equation are doubled, then K for the new equation is also double of the original K .

Explanation

To solve the question regarding the effect of doubling the coefficients of a reaction on the equilibrium constant (K), we can follow these steps:

### Step 1: Write the original reaction
Consider a general reaction:
\[ aA \rightleftharpoons bB \]
where \( a \) and \( b \) are the coefficients of reactants \( A \) and products \( B \), respectively.

### Step 2: Write the expression for the equilibrium constant (K)
The equilibrium constant \( K \) for this reaction is given by:
\[ K = \frac{[B]^b}{[A]^a} \]
where \([B]\) and \([A]\) are the molar concentrations of products and reactants at equilibrium.

### Step 3: Double the coefficients in the reaction
If we double the coefficients, the reaction becomes:
\[ 2A \rightleftharpoons 2B \]

### Step 4: Write the new expression for the equilibrium constant (K')
For the new reaction, the equilibrium constant \( K' \) is:
\[ K' = \frac{[B]^2}{[A]^2} \]

### Step 5: Relate K' to K
Notice that:
\[ K' = \frac{[B]^2}{[A]^2} = \left(\frac{[B]}{[A]}\right)^2 = K^2 \]
This shows that the new equilibrium constant \( K' \) is equal to \( K^2 \), not double \( K \).

### Conclusion
The statement that if the coefficients of a given reaction equation are doubled, then \( K \) for the new equation is also double the original \( K \) is **false**. Instead, the new equilibrium constant \( K' \) is equal to \( K^2 \).