If the initial pressure of NH3(g) is 7.845 atm, calculate the % of NH3(g) left over after the reaction reaches equilibrium according to the balanced equation. The value of Kp at 600.0 °C is 455000.00. The initial pressure of the reaction products is 0 atm.
2NH3(g) = N2(g)+3H2(g)
is there another way to approach this problem
Another way than what? The ICE way is the only one I know. I assume you know that way, too.
Yes, there is another way to approach this problem using the concept of reaction quotient (Q).
The reaction quotient (Q) is calculated in the same way as the equilibrium constant (K), but it is calculated at any point during the reaction, not just at equilibrium.
The general expression for the reaction quotient (Q) is:
Q = (partial pressure of N2)^a * (partial pressure of H2)^b / (partial pressure of NH3)^c,
where a, b, and c are the coefficients of the balanced equation.
In this case, the balanced equation is: 2NH3(g) = N2(g) + 3H2(g). So the reaction quotient (Q) is:
Q = (partial pressure of N2)^1 * (partial pressure of H2)^3 / (partial pressure of NH3)^2.
At equilibrium, Q = Kp. So we can set up the equation:
Kp = (partial pressure of N2)^1 * (partial pressure of H2)^3 / (partial pressure of NH3)^2.
Now let's substitute the given values in the equation and solve for the partial pressure of NH3 at equilibrium.
Kp = 455000.00 (given)
Partial pressure of N2 = 0 (given)
Partial pressure of H2 = 0 (given)
Let's denote the percent of NH3 left over at equilibrium as x. Then the partial pressure of NH3 at equilibrium is 7.845 - (7.845 * x/100).
Substituting these values in the equation, we get:
455000.00 = 0^1 * 0^3 / (7.845 - (7.845 * x/100))^2.
Simplifying the equation, we have:
455000.00 = 0 / (7.845 - 0.07845x)^2.
Since the partial pressure of NH3 is positive, we can ignore the solution where x is greater than 100 (as it would make the partial pressure of NH3 negative).
Solving the equation for x, we find that x is approximately 92.1%.
Therefore, the percent of NH3 left over after the reaction reaches equilibrium is approximately 92.1%.