Value of K_p for the equilibrium reaction N2O4(g) 2NO2(g) at 288 K is 47.9. The K_c for this rea

Q: Why do we use $R = 0.083$ instead of the usual $8.314$?

The value of the universal gas constant $R$ depends entirely on the units being used. Since the problem involves $K_p$ (which is based on partial pressures in bar) and $K_c$ (based on concentrations in mol/L), we must use $R = 0.083\text{ L bar K}^{-1}\text{mol}^{-1}$ to ensure all units cancel out correctly.

Q: What happens if $\Delta n_g$ turns out to be negative?

The formula $K_p = K_c(RT)^{\Delta n_g}$ still works perfectly! If $\Delta n_g$ is negative, it simply means that $K_p$ will be smaller than $K_c$, because you are effectively dividing $K_c$ by a factor of $RT$.

Q: Does the volume of the container affect the values of $K_p$ or $K_c$?

No. For a given reaction, the equilibrium constants $K_p$ and $K_c$ are dependent only on temperature. While changing the volume might shift the equilibrium position (Le Chatelier's Principle), the actual numerical value of the constant remains the same.

Q: Can $K_p$ and $K_c$ ever be exactly equal?

Yes! If a reaction has the same number of moles of gaseous products as gaseous reactants, then $\Delta n_g = 0$. Since any non-zero number to the power of zero is $1$, the term $(RT)^0$ becomes $1$, making $K_p = K_c$.

Q: Why do we ignore solids and liquids when calculating $\Delta n_g$?

The relationship between $K_p$ and $K_c$ is derived from the Ideal Gas Law ($PV=nRT$). Solids and liquids do not behave like gases; their concentrations (density) remain essentially constant and are incorporated into the equilibrium constant itself. Therefore, they do not exert partial pressures in the same way gases do.

Question

Value of $K_p$ for the equilibrium reaction
$N_2O_4(g) ightleftharpoons 2NO_2(g)$
at $288	ext{ K}$ is $47.9$.
The $K_c$ for this reaction at same temperature is ...... (Nearest integer)
($R = 0.083	ext{ L bar K}^{-1}	ext{mol}^{-1}$)

Sigma Percentile · Accepted Answer

### The Equilibrium Puzzle

Imagine you are observing a sealed glass container at exactly $288	ext{ K}$. Inside, a fascinating invisible dance is taking place. Colorless dinitrogen tetroxide ($N_2O_4$) gas molecules are constantly breaking apart to form reddish-brown nitrogen dioxide ($NO_2$) molecules, while simultaneously, $NO_2$ molecules are colliding and recombining to form $N_2O_4$.

When the rates of these two opposing processes become perfectly equal, the system reaches **chemical equilibrium**. For this specific reaction, $N_2O_4(g) ightleftharpoons 2NO_2(g)$, we are given the equilibrium constant in terms of partial pressures, $K_p = 47.9$. Our mission is to find its counterpart, $K_c$, which is the equilibrium constant expressed in terms of molar concentrations.

### The Master Equation

How do we bridge the gap between pressure and concentration? The secret lies in the Ideal Gas Law: $PV = nRT$. If we rearrange this to solve for pressure, we get $P = \left(\frac{n}{V}ight)RT$. Since moles per unit volume ($\frac{n}{V}$) is exactly what molar concentration ($C$) is, we can write $P = CRT$.

When we substitute this relationship into the equilibrium expressions for $K_p$ and $K_c$, we derive the master equation:

$$K_p = K_c(RT)^{\Delta n_g}$$

Here, **$\Delta n_g$** is the crucial factor. It represents the change in the number of moles of gaseous substances during the reaction.

### Calculating the Moles

Let's look closely at our balanced chemical equation:

$$N_2O_4(g) ightleftharpoons 2NO_2(g)$$

On the product side (the right), we have $2$ moles of $NO_2$ gas. On the reactant side (the left), we have $1$ mole of $N_2O_4$ gas. Therefore, the change in gaseous moles is:

$$\Delta n_g = 	ext{Moles of gaseous products} - 	ext{Moles of gaseous reactants}$$
$$\Delta n_g = 2 - 1 = 1$$

### The Final Calculation

Now that we have all our puzzle pieces, let's plug them into the master equation. We know $K_p = 47.9$, $R = 0.083	ext{ L bar K}^{-1}	ext{mol}^{-1}$, $T = 288	ext{ K}$, and $\Delta n_g = 1$.

Rearranging the formula to solve for $K_c$:

$$K_c = \frac{K_p}{(RT)^{\Delta n_g}}$$

Substituting the values:

$$K_c = \frac{47.9}{(0.083 	imes 288)^1}$$

First, let's compute the denominator. Multiplying the gas constant by the temperature gives us the thermal energy factor:

$$0.083 	imes 288 = 23.904$$

Finally, we divide our $K_p$ by this value:

$$K_c = \frac{47.9}{23.904} \approx 2.0038$$

The question asks for the nearest integer. Rounding $2.0038$ gives us our final, elegant answer: **$2$**.

Animated Solution for Chemistry - Chemical Equilibrium: Value of $K_{p}$ for the equilibrium reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ at $288 K$ is $47.9$ . The $K_{c}$ for this reaction at same temperature is ...... (Nearest integer) ( $R = 0.083 L bar K^{- 1} mol^{- 1}$ )

Enter Numerical Value:

Visualized Solution

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$

$K_{p} and K_{c} Relation$

$Δ n_{g} Calculation$

$Substitution$

$Denominator Calculation$

$Final Answer$

$Stoichiometry Check$

The Sigma Insight: Law of Mass Action

The Equilibrium Puzzle

The Master Equation

Calculating the Moles

The Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N2​O4​(g)⇌2NO2​(g)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kp​ and Kc​ Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δng​ Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Denominator Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Stoichiometry Check