Question

How do you write the ${K_{eq}}$ equation?

Answer 1

The ${K_{eq}}$ condition is composed as a numerical steady being equivalent to the result of the convergence of the products isolated by the result of the grouping of the reactants.

Complete step by step answer:
It is essential to recollect it is the product of the concentration.
In the event that there is a coefficient of at least two in the condition, at that point that concentration should be increased twice ($squared$).
The numerator is consistently the correct side of the condition the products
The denominator is consistently the left half of the condition the reactants.
$A{\text{ }} + {\text{ }}B{\text{ }} = {\text{ }}C{\text{ }} + {\text{ }}2{\text{ }}D$
{K_{eq}}{\text{ }} = \;\dfrac{{\left\\{ C \right\\} \times {{\left\\{ D \right\\}}^2}}}{{\left\\{ A \right\\} \times \left\\{ B \right\\}}}
Think about the accompanying reaction:
$jA{\text{ }} + {\text{ }}kB \leftrightarrow {\text{ }}lC{\text{ }} + {\text{ }}mD{\text{ }}$
When inferring the condition for ${K_{eq}}$, we accept that the forward and switch responses are rudimentary advances, hence, the stoichiometric coefficients can be utilized as the request for every reactant or item.
The rate for the forward response is ${v_f}{\text{ }} = {\text{ }}{k_f}{\text{ }}{\left[ A \right]^j}{\text{ }}{\left[ B \right]^k}$
The rate for the converse response is ${v_r}{\text{ }} = {\text{ }}{k_r}{\text{ }}{\left[ C \right]^l}{\text{ }}{\left[ D \right]^m}$
At balance, these two articulations are equivalent. With the end goal that: ${k_f}{\text{ }}{\left[ A \right]^j}{\text{ }}{\left[ B \right]^k}{\text{ }} = {\text{ }}{k_r}{\text{ }}{\left[ C \right]^l}{\text{ }}{\left[ D \right]^m}$
This can be reworked to: $\dfrac{{{k_f}}}{{{k_r}}}{\text{ }} = {\text{ }}\dfrac{{{{\left[ C \right]}^l}{\text{ }}{{\left[ D \right]}^m}}}{{{{\left[ A \right]}^j}{\text{ }}{{\left[ B \right]}^k}}}{\text{ }} = {\text{ }}Keq$
Balance constants are given without units. It is expected that every focus an incentive in the articulation is isolated by $1$ unit standard action or concentration. The standard unit of concentration is $mole/liter$. Accordingly, $x{\text{ }}moles/liter$ partitioned by $1{\text{ }}mole/liter$ leaves you with the worth $x$ and no focus units.
${K_{eq}}$ is a mathematical worth that is reliant on the stoichiometry coefficients. Hence, the synthetic condition that ${K_{eq}}$ alludes to should be characterized.
For instance:
$R \times {n_1}$ : ${H_2}{\text{ }} + {\text{ }}{I_2}{\text{ }} = {\text{ }}2HI$ : ${K_{eq}}{\text{ }} = {\text{ }}\dfrac{{{{\left[ {HI} \right]}^2}}}{{\left[ {{H_2}} \right]{\text{ }}\left[ {{I_2}} \right]}}$.
$R \times {n_2}$ : $2{H_2}{\text{ }} + {\text{ 2}}{I_2}{\text{ }} = {\text{ 4}}HI$ : ${K_{eq}}{\text{ }} = {\text{ }}\dfrac{{{{\left[ {HI} \right]}^4}}}{{\left[ {{H_2}} \right]{{\text{ }}^2}{\text{ }}\left[ {{I_2}} \right]{{\text{ }}^2}}}$.

Note:
When the subsequent condition was duplicated by$2$, the ${K_{eq}}$ was expanded to the subsequent force. When all is said in done: If $R \times {n_2}{\text{ }} = {\text{ }}n \times [R \times {n_1}]$, at that point the ${K_{eq}}$ will be connected by ${K_{eq2}}{\text{ }} = {\text{ }}\left( {{K_{eq1}}} \right){{\text{ }}^n}$ If the response is switched, this is equivalent to duplicating the condition by $-{\text{ }}1$ and the ${K_{eq\left( {reverse} \right)}}{\text{ }} = {\text{ }}{K_{eq\left( {forward} \right)}}{{\text{ }}^{ - {\text{ }}1}}$or $\dfrac{1}{{{K_{eq\left( {forward} \right)}}}}$.