I posted this below and received help but im still lost and worked out the problem but am getting the wring answer:
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Given the equilibrium constant values:
1.N2(g)+ 1/2O2(g)<---> N2O(g)
KC=2.7*10^{-18}
2.N2O4(g)<----> 2NO2(g)
KC= 4.6*10^{-3}
3. 1/2N2(g)+ O2(g)<----> NO2(g)
KC=4.1*10^{-9}
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2N2O(g)+ 3O2 (g) <----> 2N2O4(g)
I need help i dunt know what to do next Part:
1.2N2(g)+ O2(g)<---> 2N2O(g) (*2)
KC=5.4*10^{-18}
2.2NO2<----> N2O4(g)
KC= 1/4.6*10^{-3}
I know i flip the second equation so Kc is flipped over 1...and i multiplied the 1st equation by 2:
Now i don't know what to do i don't know how they got to the final equation.
Responses
* Chemistry....Please help - DrBob222, Sunday, February 8, 2009 at 7:24pm
Reverse 1.
Multiply 3 by 2.
Reverse 2.
Add the equations and you get 1/2 of the equation you want (if I didn't goof), then double that (and square the resulting K).
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Okay so this is what i did:
1.FLIP= N2O <--> N2 + 1/2O2
Kc= 1/2.7*10^{-18}= 3.7*10^17
2. FLIP= 2NO2 <---> 2NO2O4
Kc= 1/4.6*10^{-3}= 217.39
3. MULTIPLY BY 2
N2 + 2O2 <---> 2NO2
Kc= 4.1*10^{-9} * 2= 8*10^-9
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THE FINAL EQUATION SHOULD BE :
2N2O(g)+ 3O2 (g) <----> 2N2O4(g)
AND I'M NOT GETTING IT RIGHT AND I DON'T UNDERSTAND WHAT GETS CANCELED OUT I'M CONFUSED AND THE FINAL Kc I KEEP GETTING IS WRONG...
Here is my take on the question without any guarantees:
1. Reverse and multiply by 2
N2(g)+ 1/2O2(g)<---> N2O(g) Kc=2.7*10^-18
2N2O –> 2N2 + O2 K1 = [1 / 2.7*10^-18]^2
2. Reverse and multiply by 2
N2O4(g)<----> 2NO2(g) Kc= 4.6*10^-3
4NO2 –> 2N2O4 K2 = [1 / 4.6*10^-3]^2
3. Multiply by 4
1/2N2(g)+ O2(g)<----> NO2(g) Kc=4.1*10^-9
2N2+ 4O2<----> 4NO2 K3 = [4.1*10^-9 ]^4
Adding left sides and right sides, and doing cancellations, we get:
2N2O(g)+ 3O2 (g) <----> 2N2O4(g) K = (K1)(K2)(K3)
When I worked the problem for you a couple days ago, I arrived at an equation with 1/2 the coefficients but I squared the final K which would have been correct. However, I DID NOT square the reverse of 2 after I multiplied the equation by 2. GK is correct and I goofed. The rest of my answer procedure correct.
Thank YOu both of you i got the answer
To solve this problem and find the final equilibrium constant (Kc) for the reaction 2N2O(g) + 3O2 (g) ↔ 2N2O4(g), you can use the following steps:
1. Reverse equation 1:
N2O(g) ↔ N2(g) + 1/2O2(g)
This changes the sign of the equilibrium constant (Kc) so that KC = 1/2.7×10^(-18) = 3.7×10^(17)
2. Multiply equation 3 by 2:
1/2N2(g) + O2(g) ↔ NO2(g)
This gives you the equation N2(g) + 2O2(g) ↔ 2NO2(g)
Multiplying the equilibrium constant by 2, you get KC = 4.1×10^(-9) * 2 = 8×10^(-9)
3. Reverse equation 2:
2NO2 ↔ N2O4(g)
This changes the sign of the equilibrium constant (Kc) so that KC = 1/4.6×10^(-3) = 217.39
4. Add the equations:
When you add these three equations together, you should get the desired final equation:
2N2O(g) + 3O2 (g) ↔ 2N2O4(g)
However, you also need to account for the changes in the equilibrium constants when adding equations. In this case, because you reversed equation 1, you need to square the corresponding KC value (3.7×10^(17))^2 = 1.369×10^(35).
So the final equilibrium constant (Kc) for the reaction 2N2O(g) + 3O2 (g) ↔ 2N2O4(g) is 1.369×10^(35).