# In which reaction does kp=kc? a) NO(g)+O2(g) <-->N2O3(g) b) N2(g)+O2(g) <-->2NO(g) c) CaCO3(s) <-->CaO(s)+CO2(g) d) N2(g)+H2O(g) <-->NO(g)+H2(g) e) None

## I am thinking either b or d because moles final - moles initial = 0

Dec 14, 2015

#### Explanation:

As you know, 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
$T$ - the temperature at which the reaction takes place
$\Delta n$ - the difference between the number of moles of gas present on the products' side and the 8*number

of moles of gas** present on the reactants' side

In order for ${K}_{c}$ and ${K}_{p}$ to be equal, you need the volume to remain constant at equilibrium, that is,

the volume occupied by the reactants to be equal to the volume occupied by the products.

In order for that to happen, you need to have equal numbers of moles of gas on the reactants' side and on the

products' side.

This will of course get you

$\Delta n = 0$

and

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

Now, before looking at how many moles of gas you have present on each side of the equilibrium, you need to make

sure that the equations are balanced.

For example, the first equilibrium is actually

$4 {\text{NO"_text((g]) + "O"_text(2(g]) rightleftharpoons 2"N"_2"O}}_{\textrm{3 \left(g\right]}}$

In this case, you have $2$ moles of gas on the products' side and $5$ moles of gas on the reactants side $\to$ ${K}_{p} \ne {K}_{c}$.

For the second equilibrium, you have

${\text{N"_text(2(g]) + "O"_text(2(g]) rightleftharpoons 2"NO}}_{\textrm{\left(g\right]}}$

Since you have $2$ moles of gas on both sides of the equilibrium, you will indeed get ${K}_{p} = {K}_{c}$.

The third equilibrium cannot match this criterion, since you only have moles of gas on the products' side.

The last equilibrium looks like this

${\text{N"_text(2(g]) + 2"H"_2"O"_text((g]) rightleftharpoons 2"NO"_text((g]) + 2"H}}_{\textrm{2 \left(g\right]}}$

This time, you have $4$ moles of gas on the products' side and $3$ moles of gas on the reactants' side $\to$ ${K}_{p} \ne {K}_{c}$.

Always make sure that you're looking at a balanced chemical equation!