Using this data,
2 NO(g) + Cl2(g) == 2 NOCl(g) Kc = 3.20 X 10-3
NO2(g) == NO(g) + ½ O2(g) Kc = 3.93
calculate a value for Kc for the reaction,
NOCl (g) + ½ O2 (g) == NO2 (g) + ½ Cl2 (g)
A. 20.2
B. 2.06 X 10-4
C. 4.49
D. 4.84 X 10-3
E. 0.223
eliminated down to b or d
but going with b unless wrong
chemistry - DrBob222, Monday, March 12, 2012 at 6:46pm
I disagree with b AND d.
Use equation 1, take 1/2 of it, and reverse it. K1 for that will be 1/sqrt Kc.
Multiply k1 by equation 2 reversed so k2 = 1/Kc
Then k1*k2 = ?
Check my manipulations to be sure you get the equation in the problem.
chemistry - bob, Monday, March 12, 2012 at 9:03pm
when you reverse it do you get a negative?
-3.20x10-3/2=-1.6x10-3
1/.04=25
1/3.93=.254
25x.245
=6.125
not an answer choice?
No negatives are involved. When reversing Keq, it is k' = 1/Keq.
My instruction was that k1 = 1/sqrt Kc. You took 1/2 of it; that's to get the coefficients right when you add one equation to the other but k1 = 1/sqrt Kc. [When taking 1/2 of an equation, k1 is sqrt Kc. Then take the reciprocal when reversing it.
That's (1/sqrt 0.0032) and I get something like 18 or so. The 1/3.93 part looks ok.
That's my answer, too.
Let's go step-by-step to calculate the value of Kc for the reaction:
1. We have the equation: 2 NOCl (g) + ½ O2 (g) ⇌ NO2 (g) + ½ Cl2 (g)
2. From equation 1: 2 NO (g) + Cl2 (g) ⇌ 2 NOCl (g), we know that K1 = 3.20 x 10^-3
3. Taking half of equation 1 and reversing it, we get: 1/2 (2 NOCl (g)) ⇌ 1/2 (2 NO (g) + Cl2 (g))
4. The K1 for the reversed equation is equal to 1/sqrt(Kc). Thus, K1 for the reversed equation is 1/sqrt(3.20 x 10^-3).
5. From equation 2: NO2 (g) ⇌ NO (g) + ½ O2 (g), we know that K2 = 3.93.
6. The K2 for the equation 2 reversed is equal to 1/Kc.
7. Now, we can calculate Kc for the reaction by multiplying K1 and K2 together.
Kc = K1 * K2
Substituting the values we have:
Kc = (1/sqrt(3.20 x 10^-3)) * (1/3.93)
Calculating this, we get:
Kc ≈ 2.06 x 10^-4
Therefore, the correct answer is option B: 2.06 x 10^-4.
To calculate the value of Kc for the reaction:
NOCl(g) + ½ O2(g) ⇌ NO2(g) + ½ Cl2(g)
You are given two equilibrium constant values:
Kc1 = 3.20 x 10^-3 (from the equation 2 NO(g) + Cl2(g) ⇌ 2 NOCl(g))
Kc2 = 3.93 (from the equation NO2(g) ⇌ NO(g) + ½ O2(g))
Firstly, take half of the equation 2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) and reverse it. The new equilibrium constant, K1, for this equation will be the reciprocal of the square root of Kc1:
K1 = 1 / √Kc1 = 1 / √(3.20 x 10^-3) = 25
Next, reverse the equation NO2(g) ⇌ NO(g) + ½ O2(g). The new equilibrium constant, K2, for this equation will be the reciprocal of Kc2:
K2 = 1 / Kc2 = 1 / 3.93 = 0.254
Finally, multiply K1 and K2 together to get the value for Kc:
Kc = K1 * K2 = 25 * 0.254 = 6.35
Based on the given answer choices, the closest value is 2.06 x 10^-4 (option B).