For the following reaction in gaseous phase
CO + 1/2 O2 - CO2. Kc / Kp is
(a) (RT)^(1/2)
(b) (RT)^(

Q: Why do we only consider gaseous moles for $\Delta n_g$?

Because $K_p$ is defined strictly in terms of partial pressures, which only apply to gases. Solids and pure liquids have constant active masses and do not exert partial pressures in the same way, so they are excluded from the equilibrium expression.

Q: What happens if $\Delta n_g$ is exactly zero?

If $\Delta n_g = 0$, then $(RT)^0 = 1$. This means $K_p = K_c$. This occurs when the number of moles of gaseous reactants perfectly equals the number of moles of gaseous products, such as in the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$.

Q: Can $\Delta n_g$ be a negative number?

Absolutely! A negative $\Delta n_g$ simply means there are fewer moles of gas produced than consumed. This leads to a decrease in total pressure if the reaction occurs in a closed container at constant volume.

Q: Why did the sign of the exponent change in the final step?

This is a basic rule of exponents. When we move a term with a negative exponent from the numerator of one side to the numerator of the other side (by cross-multiplying), its exponent becomes positive. Mathematically, $x^{-a} = \frac{1}{x^a}$.

Q: Is this formula applicable to all chemical reactions?

It is applicable to any reversible reaction involving gases, assuming the gases behave ideally. If a reaction only involves aqueous solutions, we typically only use $K_c$.

Question

For the following reaction in gaseous phase
$CO + \frac{1}{2}O_2 \longrightarrow CO_2$. $K_c / K_p$ is

Options:

(A) $(RT)^{1/2}$

(B) $(RT)^{-1/2}$

(C) $(RT)$

(D) $(RT)^{-1}$

Sigma Percentile · Accepted Answer

### The Bridge Between $K_p$ and $K_c$: A Tale of Gaseous Moles

When dealing with chemical equilibrium in the gaseous phase, we often encounter two distinct equilibrium constants: $K_c$, which is based on molar concentrations, and $K_p$, which is based on partial pressures.

Understanding how to seamlessly convert between these two is a fundamental skill in physical chemistry. Let's dive into a classic problem that tests exactly this relationship.

### Analyzing the Setup

We are given the following gaseous reaction:

$$CO(g) + \frac{1}{2}O_2(g) \longrightarrow CO_2(g)$$

Our goal is to find the ratio $\frac{K_c}{K_p}$. To do this, we must recall the master equation that acts as a bridge between these two constants:

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

Here, $R$ is the universal gas constant, $T$ is the absolute temperature, and $\Delta n_g$ is the crucial term we need to evaluate.

### The Crucial Term: $\Delta n_g$

The term $\Delta n_g$ represents the change in the number of moles of **gaseous** substances during the reaction. It is calculated as the sum of the stoichiometric coefficients of the gaseous products minus the sum of the stoichiometric coefficients of the gaseous reactants.

Let's calculate it for our specific reaction. On the product side, we have exactly $1$ mole of $CO_2$ gas.

On the reactant side, we have $1$ mole of $CO$ gas and $\frac{1}{2}$ mole of $O_2$ gas, giving us a total of $1.5$ moles of gaseous reactants.

$$\Delta n_g = n_{	ext{products}} - n_{	ext{reactants}}$$
$$\Delta n_g = 1 - \left(1 + \frac{1}{2}ight)$$
$$\Delta n_g = 1 - 1.5 = -0.5 = -\frac{1}{2}$$

### Final Calculation

Now that we have our $\Delta n_g$, we can substitute it back into our master equation:

$$K_p = K_c(RT)^{-1/2}$$

To find the requested ratio $\frac{K_c}{K_p}$, we need to perform a simple algebraic rearrangement. A negative exponent indicates that the term belongs in the denominator.

$$K_p = \frac{K_c}{(RT)^{1/2}}$$

By cross-multiplying, we bring $(RT)^{1/2}$ to the left side and $K_p$ to the right side's denominator:

**$$\frac{K_c}{K_p} = (RT)^{1/2}$$**

And there we have it! The elegant relationship between the two equilibrium constants for this specific reaction.

Animated Solution for Chemistry - Chemical Equilibrium: For the following reaction in gaseous phase $C O + \frac{1}{2} O_{2} ⟶ C O_{2}$ . $K_{c} / K_{p}$ is

Select Answer:

Visualized Solution

$Analyzing the Reaction$

$The Master Equation$

$Calculating Δ n_{g}$

$Substitution$

$Final Rearrangement$

$The Way Forward$

The Sigma Insight: Law of Mass Action

The Bridge Between $K_{p}$ and $K_{c}$

Analyzing the Setup

The Crucial Term: $Δ n_{g}$

Final Calculation

Similar Questions

For the reaction, $C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g)$ , the $K_{p} / K_{c}$ is equal to

For the reaction, $2 NO_{2} (g) ⇌ 2 NO (g) + O_{2} (g)$ $[K_{c} = 1.8 \times 1 0^{- 6} at 18 4^{\circ} C, R = 0.00831 kJ / (mol K)]$ When $K_{p}$ and $K_{c}$ are compared at $18 4^{\circ} C$ it is found that

A vessel at $1000 K$ contains $CO_{2}$ with a pressure of $0.5 atm$ . Some of the $CO_{2}$ is converted into $CO$ on the addition of graphite. If the total pressure at equilibrium is $0.8 atm$ , the value of $K_{p}$ is

In which one of the following equilibria, $K_{p} \neq = K_{c}$ ?

For the reaction, $Fe_{2} N (s) + \frac{3}{2} H_{2} (g) ⇌ 2 Fe (s) + NH_{3} (g)$

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) ( $R = 0.083 L bar K^{- 1} mol^{- 1}$ )

The values of $\frac{K _{p}}{K _{C}}$ for the following reactions at $300 K$ are, respectively (At $300 K$ , $R T = 24.62 dm^{3} atm mol^{- 1}$ ) $N_{2} (g) + O_{2} (g) ⇌ 2 NO (g)$ $N_{2} O_{4} (g) ⇌ 2 NO_{2} (g)$ $N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)$

The equilibrium constant ( $K_{c}$ ) for the reaction $N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$ at temperature $T$ is $4 \times 1 0^{- 4}$ . The value of $K_{c}$ for the reaction $N O (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g)$ at the same temperature is

For the reaction equilibrium, $N_{2} O_{4} (g) ⇌ 2 NO_{2} (g)$ the concentrations of $N_{2} O_{4}$ and $NO_{2}$ at equilibrium are $4.8 \times 1 0^{- 2}$ and $1.2 \times 1 0^{- 2} mol L^{- 1}$ , respectively. The value of $K_{c}$ for the reaction is

Animated Solution for Chemistry - Chemical Equilibrium: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the following reaction in gaseous phase CO+21​O2​⟶CO2​. Kc​/Kp​ is

Select Answer:

Visualized Solution

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

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

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

.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; } Final Rearrangement

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