Question

At T(K), $K_p$ value for the reaction, $2\text{AO}_2(\text{g})+\text{O}_2(\text{g}) \rightleftharpoons 2\text{AO}_3(\text{g})$ is $4\times 10^{10}$. What is the $K'_p$ value for $3\text{AO}_2(\text{g})+\frac{3}{2}\text{O}_2(\text{g}) \rightleftharpoons 3\text{AO}_3(\text{g})$ at T(K)?

• A. $16\times 10^{20}$
• B. $8\times 10^{20}$
• C. $16\times 10^{15}$
• D. $8\times 10^{15}$

Hint:

For chemical equilibria, remember how the equilibrium constant is affected by algebraic manipulation of the reaction equation: - Reverse the reaction: $K_{rev} = 1/K$. - Multiply by a factor $n$: $K_{new} = K^n$. - Add two reactions: $K_{new} = K_1 K_2$.

Answer 1

The Correct Option is D

Step 1: State the relationship for equilibrium constants under scaling.
If a reaction has an equilibrium constant $K$, and the reaction is multiplied by a scaling factor $n$, the new equilibrium constant $K'$ for the scaled reaction is: \[ K' = K^n. \]

Step 2: Identify the original reaction and the new reaction.
Original reaction ($R$): $2\text{AO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{AO}_3(\text{g})$. The equilibrium constant is $K_p = 4 \times 10^{10}$. New reaction ($R'$): $3\text{AO}_2(\text{g}) + \frac{3}{2}\text{O}_2(\text{g}) \rightleftharpoons 3\text{AO}_3(\text{g})$. The equilibrium constant is $K'_p$.

Step 3: Determine the scaling factor $n$.
We find the ratio of the stoichiometric coefficients of the new reaction ($R'$) to the old reaction ($R$). For $\text{AO}_2$: $\frac{3}{2} = 1.5$. For $\text{O}_2$: $\frac{3/2}{1} = 1.5$. For $\text{AO}_3$: $\frac{3}{2} = 1.5$. The scaling factor is $n = 1.5 = \frac{3}{2}$.

Step 4: Calculate the new equilibrium constant $K'_p$.
\[ K'_p = K_p^n = (4 \times 10^{10})^{3/2}. \] \[ K'_p = (4)^{3/2} \times (10^{10})^{3/2}. \]

Step 5: Calculate the final numerical value.
Calculate the first part: $(4)^{3/2} = (\sqrt{4})^3 = 2^3 = 8$. Calculate the second part: $(10^{10})^{3/2} = 10^{(10 \times 3/2)} = 10^{15}$. \[ K'_p = 8 \times 10^{15}. \]