The value of K_C is 64 at 800 K for the reaction, N2(g) + 3H2(g) 2NH3(g) The value of K_C for th

Q: Why does the equilibrium constant become its reciprocal when we reverse the reaction?

The equilibrium constant is defined as the ratio of the concentrations of products to reactants, each raised to their stoichiometric coefficients. When you reverse a reaction, what were once products are now reactants, and vice versa. This flips the numerator and denominator in the equilibrium expression, resulting in the reciprocal of the original constant.

Q: Why do we raise the equilibrium constant to a power when we multiply the reaction by a number?

In the equilibrium expression, the stoichiometric coefficients of the balanced chemical equation become the exponents of the concentrations. If you multiply the entire equation by a factor $n$, every coefficient is multiplied by $n$. By the laws of exponents, this is mathematically equivalent to raising the entire original equilibrium expression to the power of $n$.

Q: Can I first multiply the reaction by $\frac{1}{2}$ and then reverse it? Will I get the same answer?

Absolutely! The order of operations does not matter. If you first multiply by $\frac{1}{2}$, the new constant is $64^{\frac{1}{2}} = 8$. Then, when you reverse the reaction, you take the reciprocal, which gives you $\frac{1}{8}$. Both paths lead to the exact same correct result.

Q: Does the temperature of 800 K affect our calculation here?

In this specific problem, the temperature is just contextual information. The equilibrium constant is indeed temperature-dependent, meaning $K_C = 64$ is only true at 800 K. However, since both the initial and target reactions are at the same temperature, we don't need to use the temperature value in our mathematical manipulations.

Q: What if the target reaction had different substances added to it?

If the target reaction involved adding two completely different reactions together, you would multiply their respective equilibrium constants. The rule is: when you add chemical equations, you multiply their $K$ values. But here, we only manipulated a single reaction, so we only used the reciprocal and power rules.

Question

The value of $K_C$ is $64$ at $800	ext{ K}$ for the reaction,
$N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$
The value of $K_C$ for the following reaction is
$NH_3(g) ightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)$

Options:

(A) $\frac{1}{64}$

(B) $8$

(C) $\frac{1}{4}$

(D) $\frac{1}{8}$

Sigma Percentile · Accepted Answer

### The Delicate Dance of Chemical Equilibrium

Imagine a bustling dance floor where couples are constantly forming and breaking apart. This is the essence of chemical equilibrium.

It is a dynamic state where the rate of the forward reaction perfectly matches the rate of the reverse reaction.

In our problem, we are looking at a classic example: the synthesis of ammonia via the Haber process.

$$
N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)
$$

We are given that at a specific temperature of $800	ext{ K}$, the equilibrium constant $K_C$ for this reaction is $64$.

This number, $64$, is our anchor. It tells us that at this temperature, the system strongly favors the formation of the product, ammonia.

But what happens when we look at this dance from a different perspective?

### The Art of Reversing Reactions

In chemistry, we can manipulate equations just like algebraic expressions, but with specific rules for the equilibrium constant.

Our target reaction has ammonia on the reactant side, which means we need to reverse our initial equation.

$$
2NH_3(g) ightleftharpoons N_2(g) + 3H_2(g)
$$

When we reverse a chemical reaction, the products become reactants and the reactants become products.

Consequently, the numerator and denominator in the equilibrium constant expression swap places.

**This means the new equilibrium constant is the exact reciprocal of the original one.**

Let's call our new constant $K_C'$. Mathematically, we can express this as:

$$
K_C' = \frac{1}{K_C}
$$

Since our original $K_C$ was $64$, our reversed reaction has an equilibrium constant of $\frac{1}{64}$.

### Scaling the Equation: The Power Rule

We are getting closer, but our reversed equation still doesn't perfectly match the target equation.

Our reversed equation shows the decomposition of $2$ moles of ammonia, but the target equation only involves $1$ mole.

To bridge this gap, we must multiply the entire reversed equation by a factor of $\frac{1}{2}$.

$$
NH_3(g) ightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)
$$

Here is where the second golden rule of equilibrium constants comes into play.

**When you multiply a chemical equation by a constant factor, the new equilibrium constant is the previous one raised to the power of that factor.**

Because the stoichiometric coefficients become exponents in the equilibrium expression, scaling the coefficients scales the exponents.

Let's call our final constant $K_C''$. We must raise our intermediate constant $K_C'$ to the power of $\frac{1}{2}$.

$$
K_C'' = (K_C')^{\frac{1}{2}}
$$

### Bringing It All Together

Now, we have our complete logical framework. It is time to execute the final calculation.

We substitute the value of $K_C'$ into our expression.

$$
K_C'' = \left(\frac{1}{64}ight)^{\frac{1}{2}}
$$

Do not let the fractional exponent intimidate you. Raising a value to the power of $\frac{1}{2}$ is mathematically identical to taking its square root.

$$
K_C'' = \frac{1}{\sqrt{64}}
$$

The square root of $1$ is simply $1$, and the square root of $64$ is exactly $8$.

Therefore, we arrive at our final, elegant result.

$$
K_C'' = \frac{1}{8}
$$

**The equilibrium constant for the target reaction is $\frac{1}{8}$.**

By systematically applying the rules of equilibrium—first reversing the reaction and then scaling it—we transformed a seemingly complex problem into a straightforward sequence of logical steps.

Mastering these fundamental properties of the equilibrium constant will give you a massive advantage in physical chemistry!

Animated Solution for Chemistry - Chemical Equilibrium: The value of $K_{C}$ is $64$ at $800 K$ for the reaction, $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ The value of $K_{C}$ for the following reaction is $N H_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$

Select Answer:

Visualized Solution

Initial Equilibrium

Reversing the Reaction

Multiplying by $\frac{1}{2}$

Calculating Final $K_{C}$

Final Answer

The Sigma Insight: Law of Mass Action

The Delicate Dance of Chemical Equilibrium

The Art of Reversing Reactions

Scaling the Equation

Bringing It All Together

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