# When 46g of I2 and 1g of H2 are heated to equilibrium at 450°C ,the equilibrium mixture contains 1.9g I2. How many moles of each gas present at equilibrium?Determine Kc and Kp for this reaction at the same temperature?

Dec 19, 2015

Here's what I got.

#### Explanation:

Start by writing the balanced chemical equation for this equilibrium reaction

${\text{I"_text(2(g]) + "H"_text(2(g]) rightleftharpoons color(red)(2)"HI}}_{\textrm{2 \left(g\right]}}$

Notice that you have a $1 : 1$ mole ratio between the two reactants and a $1 : \textcolor{red}{2}$ mole ratio between the reactants and the product.

Use the molar masses of hydrogen gas and iodine to determine how many moles of each you're adding to the container

46 color(red)(cancel(color(black)("g"))) * "1 mole I"_2/(253.81color(red)(cancel(color(black)("g")))) = "0.1812 moles I"_2

and

1 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.0159color(red)(cancel(color(black)("g")))) = "0.4961 moles H"_2

You know that at equilibrium, the reaction vessel contains $\text{1.9 g}$ of iodine. Calculate how many moles of iodine you have present at equilibrium

1.9 color(red)(cancel(color(black)("g"))) * "1 mole I"_2/(253.81 color(red)(cancel(color(black)("g")))) = "0.007486 moles I"_2

This tells you that a total of

${n}_{\text{converted" = 0.1812 - 0.007486 = "0.1737 moles I}} _ 2$

have been converted to hydrogen iodide, $\text{HI}$.
According to the $1 : \textcolor{red}{2}$ mole ratio that exists between iodine and hydrogen iodide, the reaction produced

0.1737 color(red)(cancel(color(black)("moles I"_2))) * (color(red)(2)" moles HI")/(1color(red)(cancel(color(black)("mole I"_2)))) = "0.3474 moles HI"

The $1 : 1$ mole ratio that exists between hydrogen gas and iodine tells you that the reaction consumed equal numbers of moles of the two reactants, so at equilibrium you'll be left with

${n}_{{H}_{2}} = 0.4961 - 0.1737 = {\text{0.3224 moles H}}_{2}$

Therefore, at equilibrium, the reaction vessel will contain - I'll leave the answers rounded to two sig figs, just for good measure

${n}_{{I}_{2}} = \textcolor{g r e e n}{{\text{0.0075 moles I}}_{2}}$

${n}_{{H}_{2}} = \textcolor{g r e e n}{{\text{0.32 moles H}}_{2}}$

${n}_{H I} = \textcolor{g r e e n}{\text{0.35 moles HI}}$

By definition, the equilibrium constant for this reaction, ${K}_{c}$, which uses equilibrium concentrations, is equal to

${K}_{c} = \left(\left[{\text{HI"]^color(red)(2))/(["I"_2] * ["H}}_{2}\right]\right)$

As you know, molarity is defined as moles of solute per liters of solution. In this case, the volume of the reaction vessel, let's say $V$, is the same for all three chemical species, so you can say that

["HI"] = "0.35 moles"/V

["H"_2] = "0.32 moles"/V

["I"_2] = "0.0075 moles"/V

This means that you have

${K}_{c} = {\left(0.35\right)}^{2} / \textcolor{red}{\cancel{\textcolor{b l a c k}{{V}^{2}}}} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}}{0.32} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}}{0.0075} = \textcolor{g r e e n}{51}$

The relationship between ${K}_{c}$ and ${K}_{p}$ is given by the equation

$\textcolor{b l u e}{{K}_{p} = {K}_{c} \cdot {\left(R T\right)}^{\Delta n}} \text{ }$, where

$R$ - the universal gas constant, equal to $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the temperature at which the reaction takes place - expressed in Kelvin
$\Delta n$ - the difference between the number of moles of gas found on the products' side and the number of moles of gas found on the reactants' side

Notice that you reaction has a total of $\textcolor{red}{2}$ moles of gas on the products' side, and $2$ moles of gas, one from each reactant, on the reactants' side.

This means that $\Delta n = 0$, which implies that

${K}_{p} = {K}_{c} \cdot {\left(R T\right)}^{0}$

${K}_{p} = {K}_{c}$