If the initial pressure of NH3(g) is 0.7317 atm, calculate the % of NH3(g) left over after the reaction reaches equilibrium according to the balanced equation. The value of Kp at 500.0 °C is 67100.00. The initial pressure of the reaction products is 0 atm.
2NH3(g) = N2(g)+3H2(g)
initial pressure NH3 = 0.7317 atm.
initial pressure H2 = 0 and N2 = 0
At equilibrium
N2 = x
H2 = 3x
NH3 = 0.7317 - 2x
Kp = 67,100 = pN2*pH2^3/pNH3^2
Substitute into Kp and solve for x, then convert to NH3, H2, N2.
Finally, pNH3 remaining (from above) to percent of original amount present.
To calculate the % of NH3(g) left over after the reaction reaches equilibrium, you first need to determine the equilibrium partial pressure of NH3 (P NH3). The equation provides the stoichiometric ratio between NH3 and the products N2 and H2.
Let's denote the initial pressure of NH3 as P initial and the equilibrium pressure of NH3 as P equilibrium.
Initially:
P NH3(initial) = 0.7317 atm
At equilibrium, the balanced equation tells us that for every 2 moles of NH3 that react, 1 mole of N2 and 3 moles of H2 are produced.
Using the information from the equation, we can set up an expression for the equilibrium pressure of NH3 (P NH3):
P NH3 = (2/2 + 1 + 3) x P initial = (6/2) x P initial = 3 x P initial
Plugging in the value of P initial and simplifying, we get:
P NH3 = 3 x 0.7317 atm = 2.1951 atm
Now that we have the equilibrium pressure of NH3, we can calculate the % of NH3 left over using the following formula:
% NH3 left over = (P NH3 / P initial) x 100%
% NH3 left over = (2.1951 atm / 0.7317 atm) x 100% = 300%
Therefore, the % of NH3 left over after the reaction reaches equilibrium is 300%.