Consider the reaction:
SO2 (g) + NO2 (g) SO3 (g) + NO (g)
At T = 1000 K, where the reaction is exothermic with an equilibrium constant K = 9.00
If the reaction vessel is instead charged initially with SO3(g) and NO(g), each at a partial pressure of 0.500 atm, the partial pressure of SO2(g) at equilibrium will be:
a) 0.050 atm
b) 0.125 atm
c) 0.250 atm
d) 0.375 atm
e) none of these
To solve this problem, we will use the ideal gas law and the equation for the equilibrium constant (K) to determine the partial pressure of SO2 at equilibrium.
Let's assume that at equilibrium, the partial pressures of SO3, NO2, SO2, and NO are P(SO3), P(NO2), P(SO2), and P(NO), respectively.
According to the balanced equation:
SO2 (g) + NO2 (g) -> SO3 (g) + NO (g)
The equation for the equilibrium constant is:
K = (P(SO3) * P(NO)) / (P(SO2) * P(NO2))
Given that K = 9.00, and initially, the partial pressures of SO3 and NO are both 0.500 atm, we can substitute these values into the equation:
9.00 = (0.500 * 0.500) / (P(SO2) * P(NO2))
Now, we can solve for P(SO2) by rearranging the equation:
P(SO2) = (0.500 * 0.500) / (9.00 * P(NO2))
Since the mole ratio of SO2 to NO2 in the balanced equation is 1:1, we know that P(SO2) = P(NO2), so we can substitute this into the equation:
P(SO2) = (0.500 * 0.500) / (9.00 * P(SO2))
Simplifying the equation further, we get:
P(SO2)^2 = (0.500 * 0.500) / (9.00)
P(SO2)^2 = 0.025 / 9.00
P(SO2)^2 = 0.002778
Taking the square root of both sides, we get:
P(SO2) = √0.002778
P(SO2) ≈ 0.0527
Therefore, the partial pressure of SO2 at equilibrium is approximately 0.0527 atm.
Since none of the given answer choices match this value exactly, the correct answer would be e) none of these.
To determine the partial pressure of SO2(g) at equilibrium, we need to use the equilibrium constant (K) and apply it to the given initial conditions.
First, we need to write the balanced equation for the reaction:
SO2(g) + NO2(g) ⇌ SO3(g) + NO(g)
The equilibrium expression can be written as:
K = [SO3] * [NO] / [SO2] * [NO2]
Given that K = 9.00, we can plug in the initial conditions as follows:
[SO3] = 0.500 atm
[NO] = 0.500 atm
[SO2] = ?
[NO2] = ?
Now, to obtain the equilibrium concentrations of NO2(g), we can use the fact that the reaction reaches equilibrium:
[SO3] / [SO2] = [NO] / [NO2]
Plugging in the initial concentrations:
0.500 / [SO2] = 0.500 / [NO2]
Since both initial concentrations are equal, we can assume that [SO2] = [NO2], so:
0.500 / [SO2] = 0.500 / [SO2]
We can simplify this equation to:
[SO2] = 0.500 atm
Therefore, the partial pressure of SO2(g) at equilibrium is 0.500 atm.
However, this option is not provided in the answer choices. So, the correct answer is e) none of these.