Consider the following reactions and their equilibrium constants.
(NO(g) + 0.5Br2(g) <===> NOBr(g) Kp = 5.3
2NO(g) <===> N2(g) + O2(g) Kp = 2.1 x 1030
Use these equations and their equilibrium constants to determine the equilibrium constant for the following reaction:
N2(g) + O2(g) + Br2(g) <===> 2NOBr(g) Kp = ?
the final answer should be 5.3^2 x( ( 2.1 x 10^30 )^-1) ) since you are doing the second reaction in reverse, you have to take the inverse of that reactions Kp value.
THANKS!
Consider the following equilibrium:
2NO(g) N2(g) + O2(g); Keq = 2.1 × 1030
The previous answer is wrong by the way.
Well, well, well! Looks like we have a fancy new reaction to tackle here. Let's see what we can do.
First, we have an equation that involves NO and Br2 forming NOBr. The equilibrium constant for this reaction, Kp, is 5.3. So, we can write it like this:
(NO(g) + 0.5Br2(g) <===> NOBr(g) Kp = 5.3
Now, we have a second equation that involves 2NO forming N2 and O2. The equilibrium constant for this reaction, Kp, is 2.1 x 10^30. So, we can write it like this:
2NO(g) <===> N2(g) + O2(g) Kp = 2.1 x 10^30
Now, let's try to combine these reactions to form the desired reaction:
N2(g) + O2(g) + Br2(g) <===> 2NOBr(g) Kp = ?
To do this, we can add the equations together and manipulate them to get the desired equation:
2NO(g) <===> N2(g) + O2(g) (1)
(NO(g) + 0.5Br2(g) <===> NOBr(g) (2)
When we add equation (1) and equation (2), we get:
2NO(g) + NO(g) + 0.5Br2(g) <===> N2(g) + 2O2(g) + NOBr(g)
Now, let's look at the equilibrium constant, Kp, for this new equation. We can determine it by multiplying the equilibrium constants of the individual reactions:
Kp = (Kp2) x (Kp1)^2
where Kp2 is the equilibrium constant for equation (2) and Kp1 is the equilibrium constant for equation (1).
So, plugging in the given values, we have:
Kp = (5.3) x (2.1 x 10^30)^2
And now, dear reader, it's time for some number crunching! Let me just grab my calculator... *beep boop beep*
Calculating... calculating... calculating...
Drum roll please...
The equilibrium constant for the reaction N2(g) + O2(g) + Br2(g) <===> 2NOBr(g) is approximately 9.81 x 10^60.
Phew! That's a pretty big number, isn't it? I hope it doesn't scare you too much. Remember, equilibrium constants tell us about the position of the equilibrium, so in this case, a value that large means the reaction is highly favored to the right, producing lots of NOBr.
I hope that answers your question! If you have any more, feel free to ask. Keep calm and chemistry on!
To determine the equilibrium constant, Kp, for the given reaction:
N2(g) + O2(g) + Br2(g) ⇌ 2NOBr(g)
You can use the equilibrium constants of the earlier given reactions and manipulate them algebraically.
Step 1: Start with the given equations and their equilibrium constants:
(NO(g) + 0.5Br2(g) ⇌ NOBr(g)) with Kp = 5.3
(2NO(g) ⇌ N2(g) + O2(g)) with Kp = 2.1 x 10^30
Step 2: Combine the two equations to obtain the desired equation:
Multiply the first equation by 2:
2(NO(g) + 0.5Br2(g) ⇌ 2NOBr(g))
Step 3: Add the first equation multiplied by 2 to the second equation:
2(NO(g) + 0.5Br2(g)) + (2NO(g) ⇌ N2(g) + O2(g)) ⇌ 2NOBr(g) + N2(g) + O2(g)
Step 4: Simplify the equation:
3NO(g) + Br2(g) + N2(g) + O2(g) ⇌ 2NOBr(g) + N2(g) + O2(g)
Step 5: Cancel out the common terms on both sides of the equation:
3NO(g) + Br2(g) ⇌ 2NOBr(g)
Step 6: Write the final equation with the equilibrium constant Kp:
Kp = Kp1 x Kp2
where Kp1 = 5.3 and Kp2 = 2.1 x 10^30
Kp = (5.3) x (2.1 x 10^30)
Step 7: Calculate the final equilibrium constant:
Kp = 1.113 x 10^31 (approximately)
Therefore, the equilibrium constant for the reaction N2(g) + O2(g) + Br2(g) ⇌ 2NOBr(g) is approximately 1.113 x 10^31.
Write K expression for eqn 1.
(NOBr)/(NO)(Br2)^1/2 = 5.3.
Now square that.
(NOBr)^2/(NO)^2(Br2) = 5.3^2
Multiply that equation by Keq for equation #2 which is
(NOBr)^2/(N2)(O2) = 2.1 x 10^30
The (NO)^2 cancels and you are left with Keq for the reaction you want and K for the final reaction is just 5.3^2 x 2.1 x 10^30 = ??