all reversible reactions
Kp= 3.5 X 10^4
temp: 1495K
N2(g)+3H2(g) reversible 2NH3(g)
1/2N2(g) + 3/2H2(g) reversible NH3(g)
what is the value of K at the temp.
what is the process for this?
please be detailed.
It's easier than it looks on the surface.
Let's call Kp = 3.5 x 10^4 the original Kp. The second equation is just 1/2 of the first one; therefore, new Kp = (original Kp)1/2.
If you had 2N2 + 6H2 <==> 4NH3, then
new Kp = (original Kp)2 . In other words, just raise the original Kp to whatever power the new equation is relative to the original coefficients. In these two examples, that is 1/2 and 2.
where does the temperature factor in? I raised it to 1/2 and got 1.75, but that was not the right answer.
(I forgot to mention this, but the first reaction is the original one)
Could you put a example up? I am very much a visual learner and need to see it before I can do it.
As long as the temperature doesn't change, K doesn't change. So for 1/2 the reactants then the new Kp = 3.5 x 10^4)1/2.
How did you get 1.75? I'm not surprised the key told you that was wrong.
I have square root of 3.5 x 10^4 = 187 and not 1.75.
Let me know if still have problems. (I just see how you got that number. You divided 3.5/2 = 1.75. You forgot the exponent of 10^4 I think.)
Is there a good website that has good examples for ap chemistry ranging from easy to difficult?
Here is one that I use often to direct students. Its' written very well and it's an AP site.
(Broken Link Removed)
calculate the standard enthalpy change for each of the following reactions. N2O4(g)+4H2(g)=N2(g)+4H2O(g)
To determine the value of K at a given temperature, we need to use the given equilibrium equation and the equilibrium constant expression. The equilibrium constant expression is written as follows:
K = [products] / [reactants]
Where [ ] represents the molar concentration of each species involved in the reaction.
In this case, we have two reversible reactions:
1) N2(g) + 3H2(g) ⇌ 2NH3(g)
2) 1/2N2(g) + 3/2H2(g) ⇌ NH3(g)
The equilibrium constant expression (K) for the first equation is:
K1 = [NH3]^2 / ([N2] * [H2]^3)
Similarly, the equilibrium constant expression (K) for the second equation is:
K2 = [NH3] / ([N2]^(1/2) * [H2]^(3/2))
We are given the value of Kp (the equilibrium constant at a specific pressure) for the first reaction, which is 3.5 × 10^4. However, we want to find the value of K (the equilibrium constant at a specific temperature, which is 1495K) for the second reaction.
To relate Kp and K, we use the ideal gas law:
Kp = K * (RT)^Δn
Where:
- Kp is the equilibrium constant expressed in terms of partial pressure.
- K is the equilibrium constant expressed in terms of concentration.
- R is the gas constant (8.314 J/(mol*K)).
- T is the temperature in Kelvin.
- Δn is the difference between the sum of the stoichiometric coefficients of the products and the sum of the stoichiometric coefficients of the reactants.
For the first reaction, Δn = (2 + 0) - (1 + 3) = -2.
Plugging in the values, we have:
3.5 × 10^4 = K1 * (8.314 * 1495)^(-2)
To solve for K1, rearrange the equation as follows:
K1 = (3.5 × 10^4) * (8.314 * 1495)^2
Now, we need to calculate the Δn for the second reaction.
Δn = (1 + 0) - (1/2 + 3/2) = 0
Since Δn is 0 for the second reaction, the equation simplifies further:
K2 = Kp
Therefore, the value of K at the given temperature is the same as Kp for the second reaction, which is 3.5 × 10^4.