In the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), the equilibrium concentration of PCl5 and PCl3 are 0.4 and 0.2 mole/litre respectively. If the value of Kc is 0.5, what is the concentration of Cl2 in mole/litre?
1.0
Understanding Chemical Equilibrium and Kc
The problem involves a reversible reaction at equilibrium: the dissociation of phosphorus pentachloride ($\text{PCl}_5$) into phosphorus trichloride ($\text{PCl}_3$) and chlorine gas ($\text{Cl}_2$).
The balanced chemical equation for this reaction is:
\(\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)\)
At equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of reactants and products remain constant. The relationship between these equilibrium concentrations is described by the equilibrium constant, $K_c$.
Equilibrium Constant Expression
For a general reversible reaction \(aA + bB \rightleftharpoons cC + dD\), the concentration-based equilibrium constant $K_c$ is given by:
\(K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\)
where $[X]$ denotes the molar concentration of species X at equilibrium, and a, b, c, and d are the stoichiometric coefficients from the balanced equation.
For the given reaction, \(\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)\), the stoichiometric coefficients for all species are 1. Therefore, the expression for $K_c$ is:
\(K_c = \frac{[\text{PCl}_3] [\text{Cl}_2]}{[\text{PCl}_5]}\)
Using Given Equilibrium Concentrations and Kc
We are provided with the following information at equilibrium:
- Equilibrium concentration of $\text{PCl}_5$, $[\text{PCl}_5] = 0.4 \text{ mole/litre}$
- Equilibrium concentration of $\text{PCl}_3$, $[\text{PCl}_3] = 0.2 \text{ mole/litre}$
- Value of the equilibrium constant, $K_c = 0.5$
We need to find the equilibrium concentration of $\text{Cl}_2$, which we can denote as $[\text{Cl}_2]$.
Calculating the Concentration of Cl2
We can substitute the given values into the $K_c$ expression:
\(0.5 = \frac{(0.2) \times [\text{Cl}_2]}{(0.4)}\)
Now, we can rearrange the equation to solve for $[\text{Cl}_2]$:
\(0.5 = \frac{0.2}{0.4} \times [\text{Cl}_2]\)
\(0.5 = 0.5 \times [\text{Cl}_2]\)
To find $[\text{Cl}_2]$, we divide both sides by 0.5:
\([\text{Cl}_2] = \frac{0.5}{0.5}\)
\([\text{Cl}_2] = 1.0 \text{ mole/litre}\)
Therefore, the equilibrium concentration of $\text{Cl}_2$ is 1.0 mole/litre.
Revision Table: Key Equilibrium Concepts
| Concept | Description | Example (for PCl5 dissociation) |
|---|---|---|
| Chemical Equilibrium | State where forward and reverse reaction rates are equal, and net changes in concentrations stop. | The point where $\text{PCl}_5 \to \text{PCl}_3 + \text{Cl}_2$ rate equals $\text{PCl}_3 + \text{Cl}_2 \to \text{PCl}_5$ rate. |
| Equilibrium Constant (Kc) | Ratio of product concentrations to reactant concentrations at equilibrium, each raised to their stoichiometric coefficients. It's constant at a given temperature. | \(K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}\) |
| Molar Concentration | Amount of solute (moles) per unit volume of solution (usually litres). Units: mole/litre or M. | Used to express $[\text{PCl}_5]$, $[\text{PCl}_3]$, and $[\text{Cl}_2]$ values. |
Additional Information: Related Concepts in Equilibrium
Besides $K_c$, another important equilibrium constant is $K_p$, which is based on partial pressures of gases involved in the reaction. For the reaction \(\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)\), the expression for $K_p$ is:
\(K_p = \frac{P_{\text{PCl}_3} P_{\text{Cl}_2}}{P_{\text{PCl}_5}}\)
where \(P_X\) is the partial pressure of gas X at equilibrium.
There is a relationship between $K_p$ and $K_c$ for gas-phase reactions:
\(K_p = K_c (RT)^{\Delta n_g}\)
where:
- \(R\) is the ideal gas constant (e.g., 0.0821 L atm mol$^{-1}$ K$^{-1}$)
- \(T\) is the absolute temperature in Kelvin
- \(\Delta n_g\) is the change in the number of moles of gas in the reaction (moles of gaseous products - moles of gaseous reactants)
For the reaction \(\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)\):
- Moles of gaseous products = 1 (for $\text{PCl}_3$) + 1 (for $\text{Cl}_2$) = 2
- Moles of gaseous reactants = 1 (for $\text{PCl}_5$)
- \(\Delta n_g = 2 - 1 = 1\)
So, for this specific reaction, the relationship is \(K_p = K_c (RT)^1\).
Understanding both $K_c$ and $K_p$, as well as their relationship, is crucial for solving various chemical equilibrium problems.
Important Questions from Chemical Equilibrium
The vapour pressures of water, acetone and ethanol at 293 K are 2.34 kPa, 12.36 kPa and 5.85 kPa respectively. Which of the following has the lowest boiling point?
Consider the gas phase reaction 2A(g) + 3B(g) ⇌ 2C(g) at a given temperature. When 2.0 moles of A(g) are reacted with 2.0 moles of B(g), 0.8 moles of C(g) are formed at equilibrium at a total pressure of 2.0 bar. The value of the equilibrium constant, KP of this reaction at the given temperature is closest to
The equilibrium dissociation energy of a diatomic molecule is 4.75 eV and its stretching frequency corresponds to 0.5 eV. The minimum energy required to dissociate the molecule in eV is
For the reaction N2(g) + 3H2(g) → 2NH3(g), the thermodynamic quantity which depends upon the pressure at which equilibrium is arrived at is (superscript 'o'represents standard state)
The equilibrium constant for the formation of the complex \(\rm Zn (NH_3)_4^{2+}\) at 25°C is close to:
[Given: \(\rm E^0_{Zn^{+2}/Zn}=-0.763V\) and \(\rm E^0_{Zn(NH_3)^{+2}_4/Zn,NH_3}=-1.03V\)]
