In an equilibrium mixture of HCl, Cl2 and H2, the partial pressure of H2 is 4.2 mPa and that of Cl2 is 8.3 mPa. What is the partial pressure of HCl at 500 oK given that at that temperature Kp = 4.0 x 1018 for
H2(g)+ Cl2(g)--> 2HCl(g)
*** I meant Kp = 4.0 x 10^18
If Kp = 4.0E18 = (HCl)^2/(H2)(Cl2)
Isn't this just a matter of substituting values for (H2) and (Cl2) and solving for (HCl)? Or is it more complicated than that.
yes i did that but the answer i get is weird
True you obtained a huge number for (HCl); however, with a K as large as 10^18 you know (HCl) must be very very large.
Is 8.35 x 10^9mPa the right answer?
@DrBob222: is that the answer?
To find the partial pressure of HCl at 500 K, we can use the equation for the equilibrium constant (Kp) at that temperature:
Kp = (P_HCl)^2 / (P_H2 * P_Cl2)
Given that Kp = 4.0 x 10^18 and P_H2 = 4.2 mPa (which is equivalent to 4.2 x 10^-6 atm) and P_Cl2 = 8.3 mPa (which is equivalent to 8.3 x 10^-6 atm), we can substitute these values into the equation:
4.0 x 10^18 = (P_HCl)^2 / ((4.2 x 10^-6 atm) * (8.3 x 10^-6 atm))
Next, we can simplify the equation by multiplying the pressures of H2 and Cl2, and then rearrange it to solve for P_HCl:
4.0 x 10^18 = (P_HCl)^2 / (3.486 x 10^-11 atm^2)
Multiplying both sides of the equation by (3.486 x 10^-11 atm^2), we get:
(3.486 x 10^-11 atm^2) * (4.0 x 10^18) = (P_HCl)^2
Now, we can calculate the left side of the equation:
1.3944 x 10^8 = (P_HCl)^2
Finally, take the square root of both sides to get the value of P_HCl:
P_HCl = √(1.3944 x 10^8)
P_HCl ≈ 1.1808 x 10^4 atm
Therefore, the partial pressure of HCl at 500 K is approximately 1.1808 x 10^4 atm.