Consider the reaction \[ \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}) \] A mixture of \( \mathrm{CO}, \mathrm{H}_{2} \), and \( \mathrm{CH}_{3} \mathrm{OH} \) is allowed to come to equilibrium at \( 225^{\circ} \mathrm{C} \). The measured equilibrium concentrations are \( [\mathrm{CO}]=0.489 \mathrm{M},\left[\mathrm{H}_{2}\right]=0.146 \mathrm{M} \), and \( \left[\mathrm{CH}_{3} \mathrm{OH}\right]=0.151 M \). What is the value of the equilibrium constant at this temperature? Express your answer numerically to one decimal place.

To find the equilibrium constant \( K_c \) for the reaction \[ \mathrm{CO}(\mathrm{g}) + 2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}), \] we can use the following formula: \[ K_c = \frac{[\mathrm{CH}_{3} \mathrm{OH}]}{[\mathrm{CO}][\mathrm{H}_{2}]^2}. \] Substituting the equilibrium concentrations into the equation: \[ K_c = \frac{[0.151]}{[0.489][0.146]^2}. \] Calculating the denominator: \[ [\mathrm{H}_{2}]^2 = (0.146)^2 = 0.021316. \] Now, substituting these values into the equation: \[ K_c = \frac{0.151}{0.489 \times 0.021316} \approx \frac{0.151}{0.010392} \approx 14.5. \] Rounding to one decimal place, the value of the equilibrium constant \( K_c \) at this temperature is \[ \boxed{14.5}. \]