Considering the following reaction: CO(g) + 2 H2(g) --> CH3OH(g) Suppose that the initial concentrations of the reactants are [CO] = 0.500 M and [H2] = 1.00 M. Assuming that there is no product at the beginning of the reaction and that at equilibrium [CO] = 0.15 M, what is the equilibrium constant at this new temperature?

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Considering the following reaction: CO(g) + 2 H2(g) --> CH3OH(g) Suppose that the initial concentrations of the reactants are [CO] = 0.500 M and [H2] = 1.00 M. Assuming that there is no product at the beginning of the reaction and that at equilibrium [CO] = 0.15 M, what is the equilibrium constant at this new temperature?

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Answer 1 · 2014-09-14T15:56:46

The equilibrium constant is 26.

First, write the balanced chemical equation with an ICE table.

CO(g) + 2H₂(g) ⇌ CH₃OH(g)

I/mol·L⁻¹: 0.500; 0.100; 0
C/mol·L⁻¹: - $x$ $x$ ; -2 $x$ $x$ ; + $x$ $x$
E/mol·L⁻¹: 0.500 - $x$ $x$ ; 0.100 - 2 $x$ $x$ ; $x$ $x$

At equilibrium, [CO] = 0.15 mol/L = (0.500 - $x$ $x$ ) mol/L

So $x$ $x$ = 0.500 – 0.15 = 0.35

Then $[H_{2}]$ $["H"_2]$ = (0.100 - 2 $x$ $x$ ) mol/L = (0.100 – 2×0.35) mol/L = 0.30 mol/L

and

$[{CH}_{3} OH]$ $["CH"_3"OH"]$ = $x$ $x$ mol/L = 0.35 mol/L

$K_{eq} = \frac{[{CH}_{3} OH]}{[CO] {[H_{2}]}_{2}^{2}} = \frac{0.35}{0.15 \times {0.30}^{2}}$ $K_"eq" = (["CH"_3"OH"])/(["CO"] ["H"_2]^2) = 0.35/(0.15 × 0.30^2)$ = 26

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