help please!
At 2200°C, K = 0.050 for the following reaction.
N2(g) + O2(g) 2 NO(g)
What is the partial pressure of NO in equilibrium with N2 and O2 that were placed in a flask at initial pressures of 0.75 atm and 0.30 atm, respectively?
Well, let me get my chemistry clown nose on and try to help you out!
First things first, we need to use the equilibrium constant expression to find out what's going on. For the reaction N2(g) + O2(g) ↔ 2 NO(g), the equilibrium constant (K) is given as 0.050 at 2200°C.
Now, let's calculate the partial pressure of NO (let's call it P(NO)) when N2 and O2 are initially at pressures of 0.75 atm and 0.30 atm, respectively.
To do this, we need to use the equation for the equilibrium constant (K) and plug in the initial pressures. The equation is: K = (P(NO))^2 / (P(N2) * P(O2))
Now, let's rearrange the equation to solve for P(NO): P(NO) = √(K * P(N2) * P(O2))
Plugging in the values, we get: P(NO) = √(0.050 * 0.75 * 0.30)
Calculating this expression gives us a result of approximately P(NO) = 0.122 atm.
So, the partial pressure of NO in equilibrium with N2 and O2, both initially at pressures of 0.75 atm and 0.30 atm, respectively, is approximately 0.122 atm.
And don't worry, even if chemistry can be a bit confusing, clown-nose mathematics is always a good laugh!
To find the partial pressure of NO in equilibrium, we need to use the equilibrium constant expression.
The balanced equation is:
N2(g) + O2(g) <=> 2 NO(g)
The equilibrium constant expression (K) is given as:
K = [NO]^2 / [N2] * [O2]
Given:
Initial pressure of N2 (P_N2) = 0.75 atm
Initial pressure of O2 (P_O2) = 0.30 atm
Equilibrium constant at 2200°C (K) = 0.050
To find the partial pressure of NO (P_NO) in equilibrium, follow these steps:
Step 1: Write the equilibrium constant expression using the given values:
0.050 = (P_NO^2) / (0.75 * 0.30)
Step 2: Rearrange the equation to solve for P_NO^2:
(P_NO^2) = 0.050 * (0.75 * 0.30)
Step 3: Calculate the value of P_NO^2:
(P_NO^2) = 0.01125
Step 4: Take the square root of both sides to determine the value of P_NO:
P_NO = √(0.01125)
Step 5: Calculate the value of P_NO:
P_NO ≈ 0.106 atm
Therefore, the partial pressure of NO in equilibrium with N2 and O2 that were placed in a flask at initial pressures of 0.75 atm and 0.30 atm, respectively, is approximately 0.106 atm.
To determine the partial pressure of NO in equilibrium, we can use the equilibrium expression and the given value of K.
The equilibrium expression for the given reaction is:
K = [NO]^2 / ([N2] * [O2])
Given the initial pressures of N2 and O2, we can convert them to molar concentrations using the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Since we are given the temperature, we can solve for the number of moles for each gas:
n(N2) = (P(N2) * V) / (R * T)
n(O2) = (P(O2) * V) / (R * T)
Substituting the values:
n(N2) = (0.75 atm * V) / (R * 2200K)
n(O2) = (0.30 atm * V) / (R * 2200K)
Now, we can substitute the expressions for the molar concentrations into the equilibrium expression:
K = [NO]^2 / ([N2] * [O2])
0.050 = [NO]^2 / ((0.75 atm * V) / (R * 2200K) * (0.30 atm * V) / (R * 2200K))
Simplifying:
0.050 = [NO]^2 / (0.0225 atm^2 * V^2 / (R^2 * 2200K^2))
To solve for [NO], we can rearrange the equation:
[NO]^2 = 0.050 * (0.0225 atm^2 * V^2 / (R^2 * 2200K^2))
[NO] = sqrt(0.050 * (0.0225 atm^2 * V^2 / (R^2 * 2200K^2)))
Here, R is the ideal gas constant (0.0821 L * atm / (mol * K)), and V is the volume in liters.
Calculating the value of [NO] using the given values, you can substitute them into the equation and calculate the partial pressure of NO.
Note: The volume of the flask is not given in the question, so you would need to have that information to obtain a numeric answer.