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

Question

Cameron Andrews · Accepted Answer

#K_c = 26#

Your tool of choice here will be the equation

#color(blue)(ul(color(black)(K_p = K_c * (RT)^(Deltan))))#

Here

Now, your reaction takes place at #1000^@"C"#, so start by converting the temperature from degrees Celsius to Kelvin.

#T = 1000^@"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"))#

#Deltan = - 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 = (2.4 * 10^(-3))/(0.0821 * 1273.15)^(-2) = color(darkgreen)(ul(color(black)(26)))#

The answer is rounded to two sig figs, the number of sig figs you have for #K_p#.

Xavier Avery · Answer

undefined To find $ K_c $ from $ K_p $, you can use the equation:

$$ K_p = K_c \left( RT \right)^\Delta n $$

Where:
- $ K_p $ is the equilibrium constant in terms of partial pressures
- $ K_c $ is the equilibrium constant in terms of concentrations
- $ R $ is the gas constant (0.0821 L atm/(mol K) or 8.314 J/(mol K))
- $ T $ is the temperature in Kelvin
- $ \Delta n $ is the change in moles of gas between products and reactants (products - reactants)

For this reaction, you'll need to know $ \Delta n $, which is the difference in the number of moles of gas on the product side minus the number of moles of gas on the reactant side.

Once you have $ \Delta n $, you can solve for $ K_c $ using the provided value of $ K_p $.