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Sagot :
Certainly! Let's work through this step-by-step to find the value of the equilibrium constant [tex]\( K_c \)[/tex] for the given reaction.
Given:
- Initial moles of [tex]\( N_2O \)[/tex]: 10.0 moles
- Volume of the container: [tex]\( 2.00 \, \text{dm}^3 \)[/tex]
- Moles of [tex]\( N_2O \)[/tex] remaining at equilibrium: 2.20 moles
The reaction is:
[tex]\[ 2 N_2O_{(g)} \rightleftharpoons 2 N_{2(g)} + O_{2(g)} \][/tex]
### Step 1: Calculate the change in moles of [tex]\( N_2O \)[/tex]
Initial moles of [tex]\( N_2O \)[/tex]: 10.0 moles
Equilibrium moles of [tex]\( N_2O \)[/tex]: 2.20 moles
Change in moles ([tex]\( \Delta N_2O \)[/tex]):
[tex]\[ \Delta N_2O = \text{Initial moles} - \text{Equilibrium moles} = 10.0 - 2.20 = 7.8 \, \text{moles} \][/tex]
### Step 2: Calculate moles of [tex]\( N_2 \)[/tex] and [tex]\( O_2 \)[/tex] produced
From the balanced reaction equation, we know that 2 moles of [tex]\( N_2O \)[/tex] decompose to produce 2 moles of [tex]\( N_2 \)[/tex] and 1 mole of [tex]\( O_2 \)[/tex].
Thus, moles of [tex]\( N_2 \)[/tex] produced ([tex]\( \Delta N_2 \)[/tex]) is equal to the change in moles of [tex]\( N_2O \)[/tex]:
[tex]\[ \Delta N_2 = 7.8 \, \text{moles} \][/tex]
Moles of [tex]\( O_2 \)[/tex] produced ([tex]\( \Delta O_2 \)[/tex]):
[tex]\[ \Delta O_2 = \frac{7.8}{2} = 3.9 \, \text{moles} \][/tex]
### Step 3: Calculate concentrations at equilibrium
The volume of the container is [tex]\( 2.00 \, \text{dm}^3 \)[/tex].
Concentration of [tex]\( N_2O \)[/tex] at equilibrium ([tex]\( [N_2O] \)[/tex]):
[tex]\[ [N_2O] = \frac{\text{Moles of } N_2O \text{ at equilibrium}}{\text{Volume}} = \frac{2.20}{2.00} = 1.1 \, \text{mol/dm}^3 \][/tex]
Concentration of [tex]\( N_2 \)[/tex] at equilibrium ([tex]\( [N_2] \)[/tex]):
[tex]\[ [N_2] = \frac{\Delta N_2}{\text{Volume}} = \frac{7.8}{2.00} = 3.9 \, \text{mol/dm}^3 \][/tex]
Concentration of [tex]\( O_2 \)[/tex] at equilibrium ([tex]\( [O_2] \)[/tex]):
[tex]\[ [O_2] = \frac{\Delta O_2}{\text{Volume}} = \frac{3.9}{2.00} = 1.95 \, \text{mol/dm}^3 \][/tex]
### Step 4: Write the expression for the equilibrium constant [tex]\( K_c \)[/tex]
The equilibrium constant [tex]\( K_c \)[/tex] for the reaction
[tex]\[ 2 N_2O_{(g)} \rightleftharpoons 2 N_{2(g)} + O_{2(g)} \][/tex]
is given by:
[tex]\[ K_c = \frac{[N_2]^2 [O_2]}{[N_2O]^2} \][/tex]
### Step 5: Substitute the equilibrium concentrations into the [tex]\( K_c \)[/tex] expression
[tex]\[ K_c = \frac{(3.9)^2 \times 1.95}{(1.1)^2} \][/tex]
### Step 6: Compute the value of [tex]\( K_c \)[/tex]
[tex]\[ K_c = \frac{15.21 \times 1.95}{1.21} \][/tex]
[tex]\[ K_c = \frac{29.6595}{1.21} \][/tex]
[tex]\[ K_c = 24.511983471074373 \][/tex]
Therefore, the equilibrium constant [tex]\( K_c \)[/tex] for the reaction is approximately 24.51.
Given:
- Initial moles of [tex]\( N_2O \)[/tex]: 10.0 moles
- Volume of the container: [tex]\( 2.00 \, \text{dm}^3 \)[/tex]
- Moles of [tex]\( N_2O \)[/tex] remaining at equilibrium: 2.20 moles
The reaction is:
[tex]\[ 2 N_2O_{(g)} \rightleftharpoons 2 N_{2(g)} + O_{2(g)} \][/tex]
### Step 1: Calculate the change in moles of [tex]\( N_2O \)[/tex]
Initial moles of [tex]\( N_2O \)[/tex]: 10.0 moles
Equilibrium moles of [tex]\( N_2O \)[/tex]: 2.20 moles
Change in moles ([tex]\( \Delta N_2O \)[/tex]):
[tex]\[ \Delta N_2O = \text{Initial moles} - \text{Equilibrium moles} = 10.0 - 2.20 = 7.8 \, \text{moles} \][/tex]
### Step 2: Calculate moles of [tex]\( N_2 \)[/tex] and [tex]\( O_2 \)[/tex] produced
From the balanced reaction equation, we know that 2 moles of [tex]\( N_2O \)[/tex] decompose to produce 2 moles of [tex]\( N_2 \)[/tex] and 1 mole of [tex]\( O_2 \)[/tex].
Thus, moles of [tex]\( N_2 \)[/tex] produced ([tex]\( \Delta N_2 \)[/tex]) is equal to the change in moles of [tex]\( N_2O \)[/tex]:
[tex]\[ \Delta N_2 = 7.8 \, \text{moles} \][/tex]
Moles of [tex]\( O_2 \)[/tex] produced ([tex]\( \Delta O_2 \)[/tex]):
[tex]\[ \Delta O_2 = \frac{7.8}{2} = 3.9 \, \text{moles} \][/tex]
### Step 3: Calculate concentrations at equilibrium
The volume of the container is [tex]\( 2.00 \, \text{dm}^3 \)[/tex].
Concentration of [tex]\( N_2O \)[/tex] at equilibrium ([tex]\( [N_2O] \)[/tex]):
[tex]\[ [N_2O] = \frac{\text{Moles of } N_2O \text{ at equilibrium}}{\text{Volume}} = \frac{2.20}{2.00} = 1.1 \, \text{mol/dm}^3 \][/tex]
Concentration of [tex]\( N_2 \)[/tex] at equilibrium ([tex]\( [N_2] \)[/tex]):
[tex]\[ [N_2] = \frac{\Delta N_2}{\text{Volume}} = \frac{7.8}{2.00} = 3.9 \, \text{mol/dm}^3 \][/tex]
Concentration of [tex]\( O_2 \)[/tex] at equilibrium ([tex]\( [O_2] \)[/tex]):
[tex]\[ [O_2] = \frac{\Delta O_2}{\text{Volume}} = \frac{3.9}{2.00} = 1.95 \, \text{mol/dm}^3 \][/tex]
### Step 4: Write the expression for the equilibrium constant [tex]\( K_c \)[/tex]
The equilibrium constant [tex]\( K_c \)[/tex] for the reaction
[tex]\[ 2 N_2O_{(g)} \rightleftharpoons 2 N_{2(g)} + O_{2(g)} \][/tex]
is given by:
[tex]\[ K_c = \frac{[N_2]^2 [O_2]}{[N_2O]^2} \][/tex]
### Step 5: Substitute the equilibrium concentrations into the [tex]\( K_c \)[/tex] expression
[tex]\[ K_c = \frac{(3.9)^2 \times 1.95}{(1.1)^2} \][/tex]
### Step 6: Compute the value of [tex]\( K_c \)[/tex]
[tex]\[ K_c = \frac{15.21 \times 1.95}{1.21} \][/tex]
[tex]\[ K_c = \frac{29.6595}{1.21} \][/tex]
[tex]\[ K_c = 24.511983471074373 \][/tex]
Therefore, the equilibrium constant [tex]\( K_c \)[/tex] for the reaction is approximately 24.51.
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