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

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1 Answer
Mar 10, 2018

Answer:

#K_c = 26#

Explanation:

Your tool of choice here will be the equation

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

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 ("atm" * "L")/("mol" * "K")#
  • #T# is the absolute temperature at which the reaction takes place
  • #Deltan# 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^@"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#.