I need part B$.$ Its the same question the ratio now it $1\mathrm{.}250$

$\frac{0\mathrm{.}56}{1}$

Methanol $)$ is formed from carbon monoxide $(A)$ and hydrogen $($B$)$ in the gas$-$phase reaction

$CO+2H_{2}C{H}_{3}OH$

The mole fractions of the reactive species at equilibrium satisfy the relation

$\frac{y_C}{y_A{y}_B^2}\frac{1}{P^2}=K_{eq}(T)$

where $P$ is the total pressure $($atm$),K_{eq}$ the reaction equilibrium constant $),$ and $T$ the temperature $($K$).$ The equilibrium constant $K_{eq}$ equals $10\mathrm{.}5$ at $373$ K $,$ and $2\mathrm{.}316{10}^{-4}$ at $573K.$A semilog plot of $K_{eq}($logarithmic scale$)$ versus $\frac{1}{T}($rectangular scale$)$ is approximately linear between $T=300K$ and $T=600K.$

a$.$ What is the value of $lo{g}_{10}(K_{eq}(1(atm){)}^2)$ at $T=422K?$

b$.$ Suppose you begin with a molar ratio of $C\frac{O}{H_2}=1\mathrm{.}000$ and no methanol. If the reaction proceeds to equilibrium at $422$ K and $2$ atm $,$ calculate the molar composition of the product and the fractional conversion of CO $.$

$y_{CO}$ :

$y_{H_2}$ :

$y_{C{H}_{3}OH}$ :

Fractional Conversion:

Repeat the calculation for a molar ratio of $C\frac{O}{H_2}=1\mathrm{.}250.$