Answer
$$[CO] = 8.7 \times 10^{3} \space M$$
Work Step by Step
- Calculate all the concentrations:
$$[H_2] = ( 1.17 \times 10^{-2})/(5.60) = 0.00209 M$$
$$[CH_3OH] = ( 3.46 \times 10^{-3})/(5.60) = 6.18 \times 10^{-4} M$$
1. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ CH_3OH ]}{[ H_2 ] ^{ 2 }[ CO ]}$$
2. Solve for the missing concentration:
$$ [CO] = \frac{[ CH_3OH ]}{[ H_2 ]^{ 2 }\times K_c}$$
3. Evaluate the expression:
$$ [CO] = \frac{( 6.18 \times 10^{-4} )}{( 0.00209 )^{ 2 }\times(1.6 \times 10^{-2})} $$
$$[CO] = 8.7 \times 10^{3} \space M$$
