(Solved): In which direction will the reaction proceed? Consider the reaction and its equilibrium constant at ...

Consider the reaction and its equilibrium constant at $25.0^{?}\mathrm{C}$ : $\mathrm{CO}(\mathrm{g})+2{\mathrm{H}}_2(\mathrm{g})={\mathrm{CH}}_3\mathrm{OH}(\mathrm{g}),K_{\mathrm{p}}=6.09\times 10^{?3}$ A reaction mixture contains $P(\mathrm{CO})=0.19\mathrm{atm},P\left({\mathrm{H}}_2\right)=0.19\mathrm{atm}$, and $P\left({\mathrm{CH}}_3\mathrm{OH}\right)=0.19$ atm. is the reaction mixture at equilibrium? Since $Q_{\mathrm{p}}=K_{\mathrm{p}}$, the reaction is not at equilibrium Since $Q_{\mathrm{p}}<K_{\mathrm{p}}$, the reaction is not at equilibrium Since $Q_{\mathrm{p}}=K_{\mathrm{p}}$, the reaction is at equilibrium Since $Q_{\mathrm{p}}>K_{\mathrm{p}}$, the reaction is not at equilibrium Correct To determine the progress of a reaction relative to the equilibrium state, calculate $Q_p$ : $\begin{array}{rl}{Q}_{\mathrm{P}}& =\frac{P\left({\mathrm{CH}}_3\mathrm{OH}\right)}{P(\mathrm{CO})P^2\left({\mathrm{H}}_2\right)}\\ & =\frac{(0.19)}{0.19\times (0.19{)}^2}\\ & =27.7\end{array}$ $Q_{\mathrm{p}}=27.7$ and $K_{\mathrm{p}}=6.09\times 10^{?3}$ since $Q_{\mathrm{p}}>K_{\mathrm{p}}$ the reaction is not at equilibrium If not, in which direction will the reaction proceed? Check all that apply. in the reverse direction to the left to the right in the forward direction