For the following reaction, 2SO3(g) = 2SO2(g) + O2(g), the equilibrium constant, Kp, is 1.32 at 627 degrees Celsius. What is the equilibrium constant for the reaction: SO3(g) = SO2(g)+ 1/2 O2(g)
.870
To find the equilibrium constant for the reaction: SO3(g) = SO2(g) + 1/2 O2(g), we need to use the relationship between the equilibrium constants of reactions with the same stoichiometry.
Given that the equilibrium constant (Kp) for the reaction 2SO3(g) = 2SO2(g) + O2(g) is 1.32 at 627 degrees Celsius, we can write the equation for this reaction as follows:
Kp = [SO2]^2 * [O2] / [SO3]^2
To find the equilibrium constant (K') for the reaction SO3(g) = SO2(g) + 1/2 O2(g), we need to manipulate the equation to match the stoichiometry of the new reaction.
First, we need to divide the given equation by 2:
1/2 * 2SO3(g) = 1/2 * 2SO2(g) + 1/2 * O2(g)
SO3(g) = SO2(g) + 1/2 * O2(g)
Next, we rearrange the equation to match the stoichiometry:
2 * SO3(g) = 2 * SO2(g) + O2(g)
2SO3(g) = 2SO2(g) + O2(g)
Comparing this new equation to the given equation, we see that they are identical. Therefore, the equilibrium constant for both reactions is the same.
The equilibrium constant for the reaction SO3(g) = SO2(g) + 1/2 O2(g) is also 1.32 at 627 degrees Celsius.
To find the equilibrium constant for the reaction:
SO3(g) = SO2(g)+ 1/2 O2(g),
we can use the relationship between equilibrium constants and chemical equations. In this case, we can write the given reaction as:
2SO3(g) = 2SO2(g) + O2(g).
Now, let's compare the two equations:
1) 2SO3(g) = 2SO2(g) + O2(g)
2) SO3(g) = SO2(g) + 1/2 O2(g)
Comparing these equations, we see that the second equation is half of the first equation.
Therefore, the equilibrium constant for the second reaction can be obtained by taking the square root of the equilibrium constant for the first reaction.
Mathematically, if Kp1 is the equilibrium constant for the first reaction, and Kp2 is the equilibrium constant for the second reaction, then:
Kp2 = sqrt(Kp1).
In this case, the given equilibrium constant, Kp, for the first reaction (2SO3(g) = 2SO2(g) + O2(g)) is 1.32.
Thus, the equilibrium constant for the second reaction (SO3(g) = SO2(g) + 1/2 O2(g)) would be:
Kp2 = sqrt(1.32).
Evaluating this expression, we find:
Kp2 ≈ 1.15.
Therefore, the equilibrium constant for the reaction SO3(g) = SO2(g)+ 1/2 O2(g) is approximately 1.15.
One thing that you should note quickly is that the equilibrium constant expression depends on how we write the chemical reaction.
For example, consider the following reaction
N2O5(g) < - > 2NO2(g) + 1/2O2(g) K = [NO2]2[O2]1/2/[N2O5]
However, we can just as easily multiply the whole reaction by 2- perhaps we don't like the stoichiometric coefficient of 1/2 in front of the oxygen. This gives
2N2O5(g) < - > 4NO2(g) + O2(g) K' = [NO2]4[O2]/[N2O5]2
K depends on how you write the chemical reaction! If you multiply the chemical equation by a given number (such as 2), it is the same as raising the equilibrium constant expression to that power. (K' = K2 in the above example.)