# If K_p = 2.4 * 10^(-3) for the reaction below, then what is K_c ?

Mar 10, 2018

${K}_{c} = 26$

#### Explanation:

Your tool of choice here will be the equation

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

Here

• ${K}_{p}$ is the equilibrium constant in terms of partial pressures
• ${K}_{c}$ is the equilibrium constant in terms of concentrations
• $R$ is the universal gas constant, equal to $0.0821 \quad \left(\text{atm" * "L")/("mol" * "K}\right)$
• $T$ is the absolute temperature at which the reaction takes place
• $\Delta n$ is the difference between the number of moles of gaseous products and the number of moles of gaseous reactants

Now, your reaction takes place at ${1000}^{\circ} \text{C}$, so start by converting the temperature from degrees Celsius to Kelvin.

$T = {1000}^{\circ} \text{C" + 273.15 = "1273.15 K}$

Notice that for every $1$ mole of nitrogen gas that takes part in the reaction, the reaction consumes $3$ moles of hydrogen gas and produces $2$ moles of ammonia.

This means that you have

Deltan = color(white)(overbrace(color(black)(" 2 "))^(color(blue)("moles of ammonia")) " "color(black)(-)" " overbrace(color(black)((" 1 + 3 ")))^(color(blue)("moles of reactants"))

$\Delta n = - 2$

Rearrange the equation to solve for ${K}_{c}$

K_c = K_p/((RT)^(Deltan)

Plug in your values to find--since you didn't provide any units for ${K}_{p}$, I'll do the calculation without added units!

${K}_{c} = \frac{2.4 \cdot {10}^{- 3}}{0.0821 \cdot 1273.15} ^ \left(- 2\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{26}}}$

The answer is rounded to two sig figs, the number of sig figs you have for ${K}_{p}$.