4Nh3(g) +3O2(g)=2N2(g)+6H2O(g) has kp=2.1x10^6 atm. Initial concentration of NH3=2.00 atm, and N2=1.00 atm, with all other concentrations being zero. Compute final concentration of NH3.
To compute the final concentration of NH3, we need to use the given equilibrium constant (Kp) and the initial concentrations of NH3, N2, and the other gases.
The balanced equation for the reaction is:
4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(g)
We are given the initial concentration of NH3 as 2.00 atm. Since the concentration of N2 is also given as 1.00 atm, we can assume the reaction has already reached equilibrium. All other concentrations are zero.
Let's denote the change in concentration of NH3 as "x" (in atm). After the reaction reaches equilibrium, the concentration of NH3 will be (2.00 - x) atm.
According to the balanced equation, the stoichiometric coefficient of NH3 is 4, which means for every 4 moles of NH3 that reacts, 2 moles of N2 are formed.
Using the equilibrium constant expression (Kp) for the given reaction:
Kp = (pN2^2 * pH2O^6) / (pNH3^4 * pO2^3)
Given pN2 = 1.00 atm, pNH3 = 2.00 - x atm, and all other concentrations being zero, we can simplify the equation as follows:
Kp = (1.00^2 * 0^6) / ((2.00 - x)^4 * 0^3)
Kp = 1 / (2.00 - x)^4
Since the value of Kp is given as 2.1x10^6 atm, we can substitute the values:
2.1x10^6 atm = 1 / (2.00 - x)^4
To solve for x, we need to rearrange the equation:
(2.00 - x)^4 = 1 / (2.1x10^6 atm)
Taking the fourth root of both sides:
2.00 - x = (1 / (2.1x10^6 atm))^(1/4)
2.00 - x = 0.00646
Solving for x:
x = 2.00 - 0.00646
x = 1.99354
Therefore, the final concentration of NH3 is approximately 1.99 atm.