Some CH4(g), initially at 1.000 atm, is placed in an empty reaction vessel. At equilibrium, the total pressure is 1.981 atm. Assuming the partial pressures are in atmospheres, what is Kp for the reaction below?
2 CH4(g) ⇌ C2H2(g) + 3 H2(g)
Look at the SO3 ==> SO2 + O2 problem below. This is done the same way. Post your work if you get stuck.
Here is the link.
https://www.jiskha.com/questions/1818084/some-so3-g-intially-at-1-000-atm-is-placed-in-an-empty-reaction-vessel-at-equilibrium
kevin/taylor/layla/lana/par/tina/lino/leo/tin/Leah ~
Please choose ONE name and keep it. Playing around with names won't get you any help sooner.
To find Kp for the given reaction, we need to know the partial pressures of all the species at equilibrium.
Let's assume that the partial pressure of CH4 at equilibrium is x atm. Since two moles of CH4 form one mole of C2H2, the partial pressure of C2H2 at equilibrium would be 2x atm. Similarly, since two moles of CH4 form three moles of H2, the partial pressure of H2 at equilibrium would be 3x atm.
According to Dalton's law of partial pressures, the total pressure at equilibrium is the sum of the partial pressures of all the species:
Total pressure = Partial pressure of CH4 + Partial pressure of C2H2 + Partial pressure of H2
1.981 atm = x atm + 2x atm + 3x atm
Simplifying the equation, we get:
1.981 atm = 6x atm
Dividing both sides of the equation by 6, we find:
x = 1.981 atm / 6
x = 0.33 atm
Now, we know the partial pressures of all the species at equilibrium:
Partial pressure of CH4 = 0.33 atm
Partial pressure of C2H2 = 2 * 0.33 atm = 0.66 atm
Partial pressure of H2 = 3 * 0.33 atm = 0.99 atm
Now, we can write the expression for Kp using the partial pressures:
Kp = (Partial pressure of C2H2 * Partial pressure of H2^3) / (Partial pressure of CH4^2)
Substituting the values we found:
Kp = (0.66 atm * 0.99 atm^3) / (0.33 atm^2)
Calculating this expression will give you the value of Kp for the given reaction.