Question

$CO\left(g\right)+2H_2\left(g\right)\to C{H}_{3}OH\left(g\right)$
$K_{eq}=14.5$
The initial concentrations of this reaction are listed below.
$\left[C{H}_{3}OH\right]=2.5M,\left[CO\right]=1.3M,\left[H_2\right]=0.6M$
Based on these initial concentrations, which statement is true?

• A. The forward rate will be greater than the reverse rate
• B. The reaction will shift left
• C. The ratio of products to that of the reactants is greater than the equilibrium constant
• D. The reaction is in equilibrium

Answer 1

The correct option is A The forward rate will be greater than the reverse rate

Given reaction is $CO\left(g\right)+2H_2\left(g\right)\to C{H}_{3}OH\left(g\right)K_{eq}=14.5$
$\left[C{H}_{3}OH\right]=2.5M,\left[CO\right]=1.3M,\left[H_2\right]=0.6M$
The reaction quotient or equilibrium constant is given below
$Qor{K}_{eq}=\frac{\left[C{H}_{3}OH\right]}{\left[\right[CO\left]\right[H_2{]}^2}$
The reaction quotient with the begining concentrations is given by
$Q=\frac{2.5}{{0.6}^2\times 1.3}=5.3$
This shows that the ratio of product is less than the equilibrium constant. Since Q is less than $K_{eq}$. In that manner the denominator will decrease and numerator will increase causes Q to become closer to $K_{eq}$.