Answer
$$K_c = 764$$
Work Step by Step
1. Find each molarity.
$ H_2 $ : ( 1.008 $\times$ 2 )= 2.016 g/mol
- Calculate the amount of moles:
$$ 0.763 \space g \times \frac{1 \space mol}{ 2.016 \space g} = 0.378 \space mol$$
- Calculate the molarity:
$$ \frac{ 0.378 \space mol}{ 3.67 \space L} = 0.103 \space M $$
$ I_2 $ : ( 126.9 $\times$ 2 )= 253.8 g/mol
- Calculate the amount of moles:
$$ 96.9 \space g \times \frac{1 \space mol}{ 253.8 \space g} = 0.382 \space mol$$
- Calculate the molarity:
$$ \frac{ 0.382 \space mol}{ 3.67 \space L} = 0.104 \space M $$
$ HI $ : ( 1.008 $\times$ 1 )+ ( 126.9 $\times$ 1 )= 127.9 g/mol
- Calculate the amount of moles:
$$ 90.4 \space g \times \frac{1 \space mol}{ 127.9 \space g} = 0.707 \space mol$$
- Calculate the molarity:
$$ \frac{ 0.707 \space mol}{ 3.67 \space L} = 0.193 \space M $$
2. At equilibrium, these are the concentrations of each compound:
$ [ H_2 ] = 0.103 \space M - x$
$ [ I_2 ] = 0.104 \space M - x$
$ [ HI ] = 0 \space M + 2x$
3. Using the concentration of $ HI $ at equilibrium, find x:
$ 0 + 2 x = 0.193 $
$ 2 x = 0.193 $
$ x = \frac{ 0.193 }{ 2 } $
$x = 0.0965 $
$ [ H_2 ] = 0.103 \space M - 0.0965 =6.50 \times 10^{-3} $
$ [ I_2 ] = 0.104 \space M - 0.0965 =7.50 \times 10^{-3} $
$ [ HI ] = 0 \space M + 2*( 0.0965 )=0.193 $
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ HI ] ^{ 2 }}{[ H_2 ][ I_2 ]}$$
4. Substitute the values and calculate the constant value:
$$K_C = \frac{( 0.193 )^{ 2 }}{( 6.50 \times 10^{-3} )( 7.50 \times 10^{-3} )} = 764$$
