The equilibrium constant (Kc) for the reaction N2(g) + O2(g) 2NO(g) at temperature T is 4 * 10^-

Q: Why does reversing the reaction invert the equilibrium constant?

The equilibrium constant is defined as the ratio of the concentration of products to reactants. When you reverse a reaction, what were once products become reactants, and vice versa. This flips the numerator and denominator in the $K_c$ expression, which is mathematically equivalent to taking the reciprocal ($1/K$).

Q: Why do we take the square root when we divide the reaction by 2?

In the equilibrium expression, the stoichiometric coefficients of the balanced equation become the exponents of the concentrations. If you multiply an entire equation by a factor $n$ (in this case, $n = 1/2$), every exponent in the $K_c$ expression is multiplied by $n$. This is equivalent to raising the entire $K_c$ value to the power of $n$. Raising a number to the power of $1/2$ is the same as taking its square root.

Q: Can an equilibrium constant ever be negative?

No, equilibrium constants can never be negative. They are calculated using concentrations or partial pressures, which are physical quantities that must always be zero or positive. Therefore, their ratio will always be positive.

Q: Does the temperature matter in this calculation?

Yes, the equilibrium constant is strictly dependent on temperature. However, the problem explicitly states that we are finding the new $K_c$ 'at the same temperature'. Because the temperature is constant, the value of $K_c$ only changes due to our mathematical manipulations of the stoichiometry, not due to thermodynamics.

Q: What would happen if we added two different reactions together?

If you add two distinct chemical reactions to form a third overall reaction, the equilibrium constant of the new overall reaction is the product of the equilibrium constants of the two original reactions ($K_{new} = K_1 \times K_2$).

Question

The equilibrium constant ($K_c$) for the reaction
$N_2(g) + O_2(g) ightleftharpoons 2NO(g)$ at temperature $T$ is $4 	imes 10^{-4}$.
The value of $K_c$ for the reaction
$NO(g) ightleftharpoons \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g)$ at the same temperature is

Options:

(A) 0.02

(B) $2.5 	imes 10^2$

(C) $4 	imes 10^{-4}$

(D) 50.0

Sigma Percentile · Accepted Answer

### The Art of Modifying Equilibrium Constants

Imagine you are an architect, and chemical reactions are your building blocks. Sometimes, the blueprint you are given isn't exactly what you need to build. You might need to flip a room upside down or scale it down to half its size. In the world of chemical equilibrium, we do this all the time, and the equilibrium constant ($K_c$) acts as our scaling factor.

Let's dive into a classic problem that tests exactly this skill. We are given a foundational reaction:

$$N_2(g) + O_2(g) ightleftharpoons 2NO(g)$$

We are told that at a specific temperature $T$, the equilibrium constant for this reaction is $K_c = 4 	imes 10^{-4}$.

Our goal is to find the equilibrium constant for a modified version of this reaction:

$$NO(g) ightleftharpoons \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g)$$

### Step 1: Reversing the Blueprint

The first thing you should notice is the position of our key player, $NO$. In the original reaction, $NO$ is a product (on the right side). However, in our target reaction, $NO$ is a reactant (on the left side).

To fix this, we must **reverse** the original reaction.

$$2NO(g) ightleftharpoons N_2(g) + O_2(g)$$

What happens to the equilibrium constant when we reverse a reaction? Think about the equilibrium expression: $K_c = \frac{[Products]}{[Reactants]}$. When we flip the reaction, the old products become the new reactants, and vice versa. Mathematically, this means we are taking the reciprocal of the original expression.

Therefore, the new constant, let's call it $K_c'$, is:

$$K_c' = \frac{1}{K_c} = \frac{1}{4 	imes 10^{-4}}$$

### Step 2: Scaling the Blueprint

We are getting closer, but we aren't there yet. Look at the coefficient of $NO$ in our reversed reaction: it is $2$. But in our target reaction, the coefficient of $NO$ is exactly $1$.

To match our target, we need to divide the entire reversed reaction by $2$ (which is the same as multiplying by $\frac{1}{2}$).

$$NO(g) ightleftharpoons \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g)$$

When we multiply a chemical equation by a stoichiometric factor $n$, the new equilibrium constant becomes the old constant raised to the power of $n$. Since we multiplied by $\frac{1}{2}$, our new constant $K_c''$ will be $K_c'$ raised to the power of $\frac{1}{2}$, which is simply the **square root**.

$$K_c'' = (K_c')^{1/2} = \sqrt{K_c'}$$

### Step 3: The Final Calculation

Now, it's just a matter of plugging in the numbers and doing the math carefully. Don't let the negative exponents intimidate you.

$$K_c'' = \sqrt{\frac{1}{4 	imes 10^{-4}}}$$

We can take the square root of the numerator and the denominator separately. The square root of $1$ is $1$. The square root of $4$ is $2$, and the square root of $10^{-4}$ is $10^{-2}$.

$$K_c'' = \frac{1}{2 	imes 10^{-2}}$$

Bringing $10^{-2}$ to the numerator makes it $10^2$, which is $100$.

$$K_c'' = \frac{100}{2} = 50$$

And there we have it! By systematically reversing and scaling our reaction, we found that the new equilibrium constant is exactly **$50.0$**. Always remember to map your target coefficients to the original ones step-by-step, and the math will naturally follow.

Select Answer:

Visualized Solution

$K_{c}$ \text{ of Original Reaction}

\text{Rules for Modifying } K

\text{Reversing the Reaction}

\text{Halving the Reaction}

\text{Final Calculation}

\text{Conclusion}

The Sigma Insight: Law of Mass Action

The Art of Modifying Equilibrium Constants

Step 1

Step 2

Step 3

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