**Practice Problem 6**

Calculate the value of *K*_{p} for the
following reaction at 500^{o}C if the value of *K*_{c}
for the reaction at this temperature is 0.040.

N_{2}(*g*) + 3 H_{2}(*g*) 2 NH_{3}(*g*)

**Solution**

The value of *K*_{p} for a reaction
can be calculated from *K*_{c} with the
following equation.

*K*_{p} = *K*_{c}
x (*RT*)^{n}

In order to use this equation, we need to know the value of *n*
for the reaction. The balanced equation for this reaction
generates two moles of products for every four moles of reactants
consumed. The value of *n* for this reaction is therefore
2 - 4, or -2.

*K*_{p} = *K*_{c}
x (*RT*)^{-2}

We now use the known value of the ideal gas constant and the
temperature in Kelvin to calculate the value of *K*_{p}
for the reaction at this temperature.

*K*_{p} = 0.040 x
[(0.08206 L-atm/mol-~~K~~)(773 ~~K~~)]^{-2}
= 1.0 x 10^{-5}