At 500 K, the equilibrium costant for the reaction `H_(2(g))+I_(2(g))hArr2HI_((g))" is "24.8" If "(1)/(2)mol//L` of HI is present at equilibrium, what are the concentrations of `H_(2)andI_(2)`, assuming that we started by taking HI and reached the equilibrium at 500 K ?

Options

A. `0.068"mol L"^(-1)`
B. `1.020"mol L"^(-1)`
C. `0.10"mol L"^(-1)`
D. `1.20"mol L"^(-1)`

Explanation

To solve the problem step by step, let's analyze the reaction and the given information:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]

### Step 2: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

### Step 3: Identify the initial concentrations
We are told that at equilibrium, the concentration of HI is \( 0.5 \, \text{mol/L} \). Assuming we started with \( 1 \, \text{mol/L} \) of HI (since we are starting with HI and reaching equilibrium), we can denote the initial concentrations as follows:
- Initial concentration of HI = \( 1.0 \, \text{mol/L} \)
- Initial concentrations of \( H_2 \) and \( I_2 \) = \( 0 \, \text{mol/L} \) (since we start with HI only)

### Step 4: Set up the change in concentrations
Let \( x \) be the amount of \( H_2 \) and \( I_2 \) that forms at equilibrium. The changes in concentration can be expressed as:
- Change in concentration of HI = \( -2x \)
- Change in concentration of \( H_2 \) = \( +x \)
- Change in concentration of \( I_2 \) = \( +x \)

At equilibrium, the concentrations will be:
- \([HI] = 1.0 - 2x\)
- \([H_2] = x\)
- \([I_2] = x\)

### Step 5: Substitute the equilibrium concentrations into the equilibrium expression
At equilibrium, we know that \([HI] = 0.5 \, \text{mol/L}\). Therefore:
\[ 0.5 = 1.0 - 2x \]
Solving for \( x \):
\[ 2x = 1.0 - 0.5 \]
\[ 2x = 0.5 \]
\[ x = 0.25 \]

### Step 6: Find the concentrations of \( H_2 \) and \( I_2 \)
Now that we have \( x \), we can find the concentrations of \( H_2 \) and \( I_2 \):
- \([H_2] = x = 0.25 \, \text{mol/L}\)
- \([I_2] = x = 0.25 \, \text{mol/L}\)

### Final Answer
The concentrations at equilibrium are:
- \([H_2] = 0.25 \, \text{mol/L}\)
- \([I_2] = 0.25 \, \text{mol/L}\)