Answer :
Certainly! Let's tackle each part of the problem step-by-step.
### (i) Forward and Reverse Reactions
The given chemical equilibrium reaction is:
[tex]\[ \text{CO}_{(g)} + 3 \text{H}_2{(g)} \rightleftharpoons \text{CH}_4{(g)} + \text{H}_2\text{O}_{(g)} \][/tex]
For writing the reactions separately:
- Forward reaction:
[tex]\[ \text{CO}_{(g)} + 3 \text{H}_2{(g)} \rightarrow \text{CH}_4{(g)} + \text{H}_2\text{O}_{(g)} \][/tex]
- Reverse reaction:
[tex]\[ \text{CH}_4{(g)} + \text{H}_2\text{O}_{(g)} \rightarrow \text{CO}_{(g)} + 3 \text{H}_2{(g)} \][/tex]
### (ii) Derive [tex]\( K_C \)[/tex] Expression for the Reaction
The equilibrium constant ([tex]\( K_C \)[/tex]) for the given reaction can be expressed in terms of the concentrations of the reactants and products.
For the reaction:
[tex]\[ \text{CO}_{(g)} + 3 \text{H}_2{(g)} \rightleftharpoons \text{CH}_4{(g)} + \text{H}_2\text{O}_{(g)} \][/tex]
The equilibrium constant [tex]\( K_C \)[/tex] is given by the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.
Thus, the expression for [tex]\( K_C \)[/tex] is:
[tex]\[ K_C = \frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3} \][/tex]
### (iii) Determine Units for [tex]\( K_C \)[/tex]
To determine the units for [tex]\( K_C \)[/tex], we need to consider the concentrations involved. Concentrations are typically expressed in moles per liter (mol/L).
In the equilibrium expression:
[tex]\[ K_C = \frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3} \][/tex]
Each concentration term has units of [tex]\( \text{mol/L} \)[/tex]. Substituting these units into the expression gives:
[tex]\[ K_C = \frac{(\text{mol/L})(\text{mol/L})}{(\text{mol/L})(\text{mol/L})^3} \][/tex]
Simplify the units:
[tex]\[ K_C = \frac{(\text{mol/L}) (\text{mol/L})}{(\text{mol/L}) (\text{mol/L}) (\text{mol/L}) (\text{mol/L})} \][/tex]
[tex]\[ K_C = \frac{(\text{mol/L})^2}{(\text{mol/L})^4} \][/tex]
[tex]\[ K_C = (\text{mol/L})^{-2} \][/tex]
So, the units for [tex]\( K_C \)[/tex] are:
[tex]\[ K_C = \text{(L/mol)}^2 \][/tex]
Thus, the units for [tex]\( K_C \)[/tex] in this reaction are [tex]\( \text{(L/mol)}^2 \)[/tex].
Now, we have written forward and reverse reactions, derived the expression for [tex]\( K_C \)[/tex], and determined its units.
### (i) Forward and Reverse Reactions
The given chemical equilibrium reaction is:
[tex]\[ \text{CO}_{(g)} + 3 \text{H}_2{(g)} \rightleftharpoons \text{CH}_4{(g)} + \text{H}_2\text{O}_{(g)} \][/tex]
For writing the reactions separately:
- Forward reaction:
[tex]\[ \text{CO}_{(g)} + 3 \text{H}_2{(g)} \rightarrow \text{CH}_4{(g)} + \text{H}_2\text{O}_{(g)} \][/tex]
- Reverse reaction:
[tex]\[ \text{CH}_4{(g)} + \text{H}_2\text{O}_{(g)} \rightarrow \text{CO}_{(g)} + 3 \text{H}_2{(g)} \][/tex]
### (ii) Derive [tex]\( K_C \)[/tex] Expression for the Reaction
The equilibrium constant ([tex]\( K_C \)[/tex]) for the given reaction can be expressed in terms of the concentrations of the reactants and products.
For the reaction:
[tex]\[ \text{CO}_{(g)} + 3 \text{H}_2{(g)} \rightleftharpoons \text{CH}_4{(g)} + \text{H}_2\text{O}_{(g)} \][/tex]
The equilibrium constant [tex]\( K_C \)[/tex] is given by the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.
Thus, the expression for [tex]\( K_C \)[/tex] is:
[tex]\[ K_C = \frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3} \][/tex]
### (iii) Determine Units for [tex]\( K_C \)[/tex]
To determine the units for [tex]\( K_C \)[/tex], we need to consider the concentrations involved. Concentrations are typically expressed in moles per liter (mol/L).
In the equilibrium expression:
[tex]\[ K_C = \frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3} \][/tex]
Each concentration term has units of [tex]\( \text{mol/L} \)[/tex]. Substituting these units into the expression gives:
[tex]\[ K_C = \frac{(\text{mol/L})(\text{mol/L})}{(\text{mol/L})(\text{mol/L})^3} \][/tex]
Simplify the units:
[tex]\[ K_C = \frac{(\text{mol/L}) (\text{mol/L})}{(\text{mol/L}) (\text{mol/L}) (\text{mol/L}) (\text{mol/L})} \][/tex]
[tex]\[ K_C = \frac{(\text{mol/L})^2}{(\text{mol/L})^4} \][/tex]
[tex]\[ K_C = (\text{mol/L})^{-2} \][/tex]
So, the units for [tex]\( K_C \)[/tex] are:
[tex]\[ K_C = \text{(L/mol)}^2 \][/tex]
Thus, the units for [tex]\( K_C \)[/tex] in this reaction are [tex]\( \text{(L/mol)}^2 \)[/tex].
Now, we have written forward and reverse reactions, derived the expression for [tex]\( K_C \)[/tex], and determined its units.
