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Step-by-Step Solution
Step 1: Identify the given reaction and its equilibrium constant
The original reaction given is:
SO3(g)
$\leftrightharpoons$
SO2(g) +
$\tfrac{1}{2}$ O2(g)
and its equilibrium constant is provided as:
$K_c = 4.9 \times 10^{-2}$
Step 2: Write the expression for $K_c$ of the given reaction
In terms of concentrations, the equilibrium constant expression for this reaction is:
$K_c = \dfrac{[SO_2] [O_2]^{\tfrac{1}{2}}}{[SO_3]} = 4.9 \times 10^{-2}$
Step 3: Relate it to the desired reaction
The desired reaction is:
2SO2(g) + O2(g)
$\leftrightharpoons$
2SO3(g)
Notice that to transform the given reaction into this new reaction, we need to square and invert appropriately because we want to relate the expressions correctly:
The new reaction is the reverse of 2 times the original reaction.
When we square the original reaction, we get an expression involving
$(SO_2)^2$, $(O_2)$, and $(SO_3)^2$.
The new reaction is also reversed; thus, we must take the reciprocal of the squared $K_c$.
Step 4: Square the $K_c$ for the doubled reaction
First, if we square the original reaction, the resultant expression is:
$\left(K_c\right)^2
= \left(4.9 \times 10^{-2}\right)^2
= 24.01 \times 10^{-4}
= 2.401 \times 10^{-3}$
(approximate value)
In terms of concentrations, squaring the original expression yields:
$\dfrac{[SO_2]^2 [O_2]}{[SO_3]^2}
= 2.401 \times 10^{-3}$
(approximately)
Step 5: Take the reciprocal for the reversed reaction
The new reaction we want is effectively the reverse of the squared original reaction, so its $K_c$ is the reciprocal of $2.401 \times 10^{-3}$. Therefore:
$K'_c = \dfrac{1}{2.401 \times 10^{-3}} \approx 416
Step 6: Final Answer
The equilibrium constant for the reaction
2SO2(g) + O2(g) $\leftrightharpoons$ 2SO3(g)
is:
$\boxed{416}$
