Question

Balance each of the following examples of heterogeneous equilibria and write each reaction quotient, \(Q_{c}\) (a) \(\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\operatorname{SnO}_{2}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \operatorname{Sn}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}(l)+\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}(l)\)

Short answer

(a) Balanced: 2 NaHCO3(s) ↔ Na2CO3(s) + CO2(g) + H2O(g); Qc=[CO2][H2O]. (b) Balanced: SnO2(s)+H2(g)↔Sn(s)+H2O(g); Qc=[H2O]/[H2]. (c) Balanced: H2SO4(l)+SO3(g)↔H2S2O7(l); Qc=1/[SO3].

Step-by-step solution

- Balance Equation (a)

First, ensure that the number of atoms of each element on the reactant side is equal to the number of atoms on the product side. The given equation is a)a)a)a)a)\[ \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g).\]Check the balancing:- Sodium (Na): 2 on the right but 1 on the left- Carbon (C): 2 on the right but 1 on the left- Oxygen (O): 5 (3 from NaHCO3) on the left and 6 (3 from Na2CO3 + 2 from CO2 + 1 from H2O) on the right- Hydrogen (H): 1 on the left but 2 on the rightThe equation is not balanced. It should be:\[ 2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\].

- Write Reaction Quotient for (a)

The reaction quotient, QC, is written from the balanced equation. For heterogeneous equilibria involving solids, we only consider the gaseous species and aqueous solutions:\[Q_{c} = \frac{[\mathrm{CO}_2][\mathrm{H}_2 \mathrm{O}]}{1} = [\mathrm{CO}_2][\mathrm{H}_2 \mathrm{O}]\].

- Balance Equation (b)

Balance the reactants and products:b)\[ \operatorname{SnO}_{2}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \operatorname{Sn}(s)+\mathrm{H}_{2} \mathrm{O}(g)\]Check the balancing:- Tin (Sn): 1 on the left and 1 on the right- Oxygen (O): 2 on the left and 1 on the right . - Hydrogen (H): 2 on either sideAlready balanced.

- Write Reaction Quotient for (b)

The reaction quotient, QC, for the balanced equation (b), considering only gaseous species:\[Q_{c} = \frac{[\mathrm{H}_2\mathrm{O}]}{[\mathrm{H}_2]}\].

- Balance Equation (c)

Balance the reactants and products:c)\[\mathrm{H}_{2} \mathrm{SO}_{4}(l)+\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}(l)\]Check the balancing:- Hydrogen (H): 2 on both sides (balanced)- Sulfur (S): 2 on both sides (balanced)- Oxygen (O): 7 on both sides (balanced)Already balanced.

- Write Reaction Quotient for (c)

The reaction quotient, QC, for the balanced equation (c), considering only gaseous species:\[Q_{c} = \frac{1}{[\mathrm{SO}_{3}]}\].

Key concepts

heterogeneous equilibria

In chemistry, equilibria are termed 'heterogeneous' when they involve more than one phase, such as solids, liquids, and gases. The key characteristic here is that the reacting species exist in different states. For instance, reaction (a) involves solids (NaHCO3 and Na2CO3) and gases (CO2 and H2O). Similarly, reaction (b) includes a solid (SnO2), a gas (H2), and another solid (Sn). The same principle applies to reaction (c), where a liquid (H2SO4), another liquid (H2S2O7), and a gas (SO3) are involved. In heterogeneous equilibria, solids and pure liquids are not included in the equilibrium expression because their concentrations do not change during the reaction. Only gases and aqueous species are included in the reaction quotient, denoted as Qc.

reaction quotient

The reaction quotient, denoted as Qc, is an important concept in chemical equilibrium. It gives an idea of the ratio of concentrations of products to reactants at any point in time. For the heterogeneous reactions provided, Qc is calculated using only the concentrations of gaseous and aqueous species.

• For reaction (a):
  \[ 2 \text{NaHCO}_3(s) \rightleftharpoons \text{Na}_2 \text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2 \text{O}(g) \] The reaction quotient, Qc, would be \[ Q_c = [\text{CO}_2][\text{H}_2 \text{O}] \]
• For reaction (b): The equation is already balanced as
  \[ \text{SnO}_2(s) + \text{H}_2(g) \rightleftharpoons \text{Sn}(s) + \text{H}_2 \text{O}(g) \] and the Qc will be
  \[ Q_c = \frac{[\text{H}_2 \text{O}]}{[\text{H}_2]} \]
• For reaction (c): The balanced reaction is
  \[ \text{H}_2 \text{SO}_4(l) + \text{SO}_3(g) \rightleftharpoons \text{H}_2 \text{~S}_2 \text{O}_7(l) \] and the Qc will be
  
  \[ Q_c = \frac{1}{[\text{SO}_3]} \]

Qc is crucial, as it helps predict which direction the reaction will proceed to reach equilibrium. If Qc < Kc (equilibrium constant), the reaction will move forward to produce more products. Conversely, if Qc > Kc, the reaction will shift backward to form more reactants.

chemical equilibrium

Chemical equilibrium refers to the state in which the concentrations of all reactants and products remain constant over time. It is a dynamic balance where the rates of the forward and reverse reactions are equal. For heterogeneous equilibria, equilibrium involves the same principles, but only the concentrations of gases and aqueous solutions change noticeably.

The equilibrium constant, Kc, is specific for a given reaction at a certain temperature. It indicates the ratio of concentrations of products to reactants at equilibrium. For a general reaction:

\[ aA + bB \rightleftharpoons cC + dD \]

The equilibrium constant expression is

\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
For the exercises given:

• (a) The equation \[2 \text{NaHCO}_3(s) \rightleftharpoons \text{Na}_2 \text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2 \text{O}(g) \]
  K_c = \[ [\text{CO}_2][\text{H}_2 \text{O}] \]
  
• (b) The equation
  \[ \text{SnO}_2(s) + \text{H}_2(g) \rightleftharpoons \text{Sn}(s) + \text{H}_2 \text{O}(g) \]
  K_c = \[ \frac{[\text{H}_2 \text{O}]}{[\text{H}_2]} \]
  
• (c) The equation
  
  \[ \text{H}_2 \text{SO}_4(l) + \text{SO}_3(g) \rightleftharpoons \text{H}_2 \text{~S}_2 \text{O}_7(l) \]
  
  K_c = \[ \frac{1}{[\text{SO}_3]} \]
  Understanding equilibrium is fundamental for predicting the behavior of chemical reactions under different conditions.