the equilibrium constant for the reaction N2 + H2 <---> 2NH3 at 700K is 12. calculate the amount of NH3 if the reaction started with 1 mole of N2 and 3 moles of H2 in a 1 liter vessel
To calculate the amount of NH3 at equilibrium, we need to use the concept of the equilibrium constant and stoichiometry.
Step 1: Write the balanced chemical equation for the reaction:
N2 + 3H2 <---> 2NH3
Step 2: Determine the initial moles of N2 and H2:
Given that we have 1 mole of N2 and 3 moles of H2 initially.
Step 3: Calculate the change in moles:
From the balanced equation, 1 mole of N2 reacts to form 2 moles of NH3. Thus, if x moles of N2 react, we will form 2x moles of NH3.
Similarly, 3 moles of H2 react to form 2 moles of NH3. So, if x moles of H2 react, we will form 2x moles of NH3.
Step 4: Write the expression for the equilibrium constant (Kc):
Kc = [NH3]^2 / [N2]·[H2]^3
Step 5: Substitute the given value of Kc and set up the equation:
12 = (2x)^2 / (1-x)·(3-3x)^3
Step 6: Solve the equation for x:
Using algebraic techniques or a numerical solver, find the value of x that satisfies the equation.
Step 7: Calculate the amount of NH3 at equilibrium:
The amount of NH3 at equilibrium will be 2x moles.
By following these steps, you can calculate the amount of NH3 at equilibrium given the initial amounts of N2 and H2 and the equilibrium constant for the reaction.