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Sagot :
Let's work through the problem step by step to find the equilibrium constant, [tex]\( K_c \)[/tex], for the reaction
[tex]\[ CH_4(g) + CO_2(g) \rightleftharpoons 2 CO (g) + 2 H_2(g) \][/tex]
### Step 1: Calculate the moles of [tex]\( H_2 \)[/tex]
We are given the mass of [tex]\( H_2 \)[/tex] as 25.3 g. First, we need to find the number of moles of [tex]\( H_2 \)[/tex].
1. The molar mass of H[tex]\(_2\)[/tex] is approximately 2 g/mol.
2. The formula to calculate the number of moles is:
[tex]\[ \text{moles of } H_2 = \frac{\text{mass of } H_2}{\text{molar mass of } H_2} \][/tex]
3. Thus:
[tex]\[ \text{moles of } H_2 = \frac{25.3 \, \text{g}}{2 \, \text{g/mol}} = 12.65 \, \text{mol} \][/tex]
### Step 2: Calculate the concentration of [tex]\( H_2 \)[/tex]
The next step is to determine the concentration of [tex]\( H_2 \)[/tex] in the flask.
1. We are given the volume of the flask as 5.00 L.
2. Concentration is calculated using the formula:
[tex]\[ \text{Concentration of } H_2 = \frac{\text{moles of } H_2}{\text{volume}} \][/tex]
3. Thus:
[tex]\[ \text{Concentration of } H_2 = \frac{12.65 \, \text{mol}}{5.00 \, \text{L}} = 2.53 \, \text{M} \][/tex]
### Step 3: Set up the expression for [tex]\( K_c \)[/tex]
The balanced chemical equation for the reaction is:
[tex]\[ CH_4(g) + CO_2(g) \rightleftharpoons 2 CO (g) + 2 H_2(g) \][/tex]
The expression for the equilibrium constant [tex]\( K_c \)[/tex] based on the given reaction is:
[tex]\[ K_c = \frac{[CO]^2 [H_2]^2}{[CH_4][CO_2]} \][/tex]
### Step 4: Determine the concentrations needed for the [tex]\( K_c \)[/tex] expression
Assuming the concentrations of [tex]\( CH_4 \)[/tex] and [tex]\( CO_2 \)[/tex] are each 1 M (as simplest assumption is made for the purpose of this problem), we need the concentration of [tex]\( CO \)[/tex] at equilibrium.
1. From the stoichiometry of the reaction, for every 2 moles of [tex]\( H_2 \)[/tex] formed, 2 moles of [tex]\( CO \)[/tex] are also formed.
2. Hence, the concentration of [tex]\( CO \)[/tex] will be half of that of [tex]\( H_2 \)[/tex] as:
[tex]\[ \text{Concentration of } CO = \frac{\text{Concentration of } H_2}{2} = \frac{2.53 \, \text{M}}{2} = 1.265 \, \text{M} \][/tex]
### Step 5: Plug the concentrations into the [tex]\( K_c \)[/tex] expression and solve
[tex]\[ K_c = \frac{(1.265 \, \text{M})^2 \cdot (2.53 \, \text{M})^2}{(1 \, \text{M}) \cdot (1 \, \text{M})} \][/tex]
Performing the calculation:
[tex]\[ K_c = \frac{(1.265)^2 \cdot (2.53)^2}{1 \cdot 1} \][/tex]
Calculate individually first:
1. [tex]\( (1.265)^2 = 1.600225 \)[/tex]
2. [tex]\( (2.53)^2 = 6.4009 \)[/tex]
So,
[tex]\[ K_c = 1.600225 \times 6.4009 = 10.2428802025 \approx 10.24 \][/tex]
### Final Answer
Thus, the equilibrium constant [tex]\( K_c \)[/tex] for the reaction is:
[tex]\[ K_c = 10.24 \][/tex]
[tex]\[ CH_4(g) + CO_2(g) \rightleftharpoons 2 CO (g) + 2 H_2(g) \][/tex]
### Step 1: Calculate the moles of [tex]\( H_2 \)[/tex]
We are given the mass of [tex]\( H_2 \)[/tex] as 25.3 g. First, we need to find the number of moles of [tex]\( H_2 \)[/tex].
1. The molar mass of H[tex]\(_2\)[/tex] is approximately 2 g/mol.
2. The formula to calculate the number of moles is:
[tex]\[ \text{moles of } H_2 = \frac{\text{mass of } H_2}{\text{molar mass of } H_2} \][/tex]
3. Thus:
[tex]\[ \text{moles of } H_2 = \frac{25.3 \, \text{g}}{2 \, \text{g/mol}} = 12.65 \, \text{mol} \][/tex]
### Step 2: Calculate the concentration of [tex]\( H_2 \)[/tex]
The next step is to determine the concentration of [tex]\( H_2 \)[/tex] in the flask.
1. We are given the volume of the flask as 5.00 L.
2. Concentration is calculated using the formula:
[tex]\[ \text{Concentration of } H_2 = \frac{\text{moles of } H_2}{\text{volume}} \][/tex]
3. Thus:
[tex]\[ \text{Concentration of } H_2 = \frac{12.65 \, \text{mol}}{5.00 \, \text{L}} = 2.53 \, \text{M} \][/tex]
### Step 3: Set up the expression for [tex]\( K_c \)[/tex]
The balanced chemical equation for the reaction is:
[tex]\[ CH_4(g) + CO_2(g) \rightleftharpoons 2 CO (g) + 2 H_2(g) \][/tex]
The expression for the equilibrium constant [tex]\( K_c \)[/tex] based on the given reaction is:
[tex]\[ K_c = \frac{[CO]^2 [H_2]^2}{[CH_4][CO_2]} \][/tex]
### Step 4: Determine the concentrations needed for the [tex]\( K_c \)[/tex] expression
Assuming the concentrations of [tex]\( CH_4 \)[/tex] and [tex]\( CO_2 \)[/tex] are each 1 M (as simplest assumption is made for the purpose of this problem), we need the concentration of [tex]\( CO \)[/tex] at equilibrium.
1. From the stoichiometry of the reaction, for every 2 moles of [tex]\( H_2 \)[/tex] formed, 2 moles of [tex]\( CO \)[/tex] are also formed.
2. Hence, the concentration of [tex]\( CO \)[/tex] will be half of that of [tex]\( H_2 \)[/tex] as:
[tex]\[ \text{Concentration of } CO = \frac{\text{Concentration of } H_2}{2} = \frac{2.53 \, \text{M}}{2} = 1.265 \, \text{M} \][/tex]
### Step 5: Plug the concentrations into the [tex]\( K_c \)[/tex] expression and solve
[tex]\[ K_c = \frac{(1.265 \, \text{M})^2 \cdot (2.53 \, \text{M})^2}{(1 \, \text{M}) \cdot (1 \, \text{M})} \][/tex]
Performing the calculation:
[tex]\[ K_c = \frac{(1.265)^2 \cdot (2.53)^2}{1 \cdot 1} \][/tex]
Calculate individually first:
1. [tex]\( (1.265)^2 = 1.600225 \)[/tex]
2. [tex]\( (2.53)^2 = 6.4009 \)[/tex]
So,
[tex]\[ K_c = 1.600225 \times 6.4009 = 10.2428802025 \approx 10.24 \][/tex]
### Final Answer
Thus, the equilibrium constant [tex]\( K_c \)[/tex] for the reaction is:
[tex]\[ K_c = 10.24 \][/tex]
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