Calculate the value of the equilibrium constant for the reaction 4 NH31g2 + 3 O21g2 ∆ 2 N21g2 + 6 H2O1g2
given the following equilibrium constants at a certain temperature.
2 H21g2 + O21g2 ∆ 2 H2O1g2 Kc = 3.2 * 1081
N21g2 + 3 H21g2 ∆ 2 NH31g2 Kc = 3.5 * 108

Question

Calculate the value of the equilibrium constant for the reaction 4 NH31g2 + 3 O21g2 ∆ 2 N21g2 + 6 H2O1g2

given the following equilibrium constants at a certain temperature.

2 H21g2 + O21g2 ∆ 2 H2O1g2 Kc = 3.2 * 1081

N21g2 + 3 H21g2 ∆ 2 NH31g2 Kc = 3.5 * 108

Accepted Answer

Welcome back, everyone. Given the following data at 500 Kelvin determine the value of the equilibrium constant at the same temperature for the reaction CH four plus 202 in equilibrium with CO2 and 2 H2O. So let's consider our target equation. And what we want to understand is that we can manipulate equilibrium one and equilibrium two in a way that we get our target equation by adding them up. Of course, we need to modify them slightly. Now, the question is how should we modify them? Well, first of all, we need to consider every species in our target equation starting with the first reactant, which is CH four. And we can notice that we have CH four in the first reaction, meaning we are not going to modify it. So let's simply rewrite it CH four plus ot in equilibrium width, carbon solid plus two H two gas. Now, we're not going to modify it because CH four is on the reactant side, it has a coefficient of one. That's exactly what we need, right? So we're going to leave it as it is. And the equilibrium constant is K one, let's skip oxygen. The reason why we're skipping oxygen is because we can see oxygen in both of our equilibria and simply move on to CO2. Now, for CO two, what we can notice is that it is on the reactant side with an equilibrium two. And we want to make sure that it is on the product side within the target equation. So what we're going to do is essentially flip it if we reverse it, that means we're multiplying all of our coefficients by negative one. Now our products become our reactants. So we get carbon solid as our first reactant reacts with 02 gas. And this is an equilibrium with co two gas. What is the equilibrium constant of the modified equation? Well, if we had K two and now we have a reverse reaction, we are simply taking the reciprocal value one divided by K two. That's the rule, right? And now essentially what we want to do is just add them up and check if we can get the target equation. On the reactant side, we have siege for gas. What about oxygens? We add them up and we get a total of two moles of oxygen. We can clearly see that carbons get canceled out because they're on opposite sides of the equation side, specifically relative to the equilibrium arrows. And now when we consider the product side, we have one mole of co two plus two moles of h2o. This is exactly what we have in the target equation. What is the equilibrium constant of this target equation? Well, if we add equations, one and two, the equilibrium constant is simply the product of the individual equilibrium constants. So what we want to do is just multiply the resultant equilibrium constants. K one multiplied by one divided by K two or simply K one divided by K two, substituting the values. Key one is seven multiplied by 10. The power of 59th divided by K two, which is nine multiplied by its sense, the power of 47th and the result is seven multiplied by sense, the power of 11th, which is the equilibrium constant of the target equation. That would be it. And thank you for watching.

Calculate the value of the equilibrium constant for the reaction 4 NH31g2 + 3 O21g2 ∆ 2 N21g2 + 6 H2O1g2
given the following equilibrium constants at a certain temperature.
2 H21g2 + O21g2 ∆ 2 H2O1g2 Kc = 3.2 * 1081
N21g2 + 3 H21g2 ∆ 2 NH31g2 Kc = 3.5 * 108

Verified Solution

Key Concepts

Equilibrium Constant (Kc)

Reaction Stoichiometry

Le Chatelier's Principle

Calculate the value of the equilibrium constant for the reaction 4 NH31g2 + 3 O21g2 ∆ 2 N21g2 + 6 H2O1g2given the following equilibrium constants at a certain temperature.2 H21g2 + O21g2 ∆ 2 H2O1g2 Kc = 3.2 * 1081N21g2 + 3 H21g2 ∆ 2 NH31g2 Kc = 3.5 * 108

Verified Solution

Key Concepts

Equilibrium Constant (Kc)

Reaction Stoichiometry

Le Chatelier's Principle

Calculate the value of the equilibrium constant for the reaction 4 NH31g2 + 3 O21g2 ∆ 2 N21g2 + 6 H2O1g2
given the following equilibrium constants at a certain temperature.
2 H21g2 + O21g2 ∆ 2 H2O1g2 Kc = 3.2 * 1081
N21g2 + 3 H21g2 ∆ 2 NH31g2 Kc = 3.5 * 108