To calculate [tex]\( K_p \)[/tex] for the reaction
[tex]\[ NO_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}, \][/tex]
we'll use the relationship between [tex]\( K_c \)[/tex] and [tex]\( K_p \)[/tex].
### Step-by-Step Solution:
1. Write the given reaction and its equilibrium constant:
The given equilibrium reaction is:
[tex]\[ N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}, \][/tex]
with [tex]\( K_c = 0.0408 \)[/tex] at 2273 K.
2. Determine the relationship between the given reaction and the target reaction:
The target reaction is:
[tex]\[ NO_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}. \][/tex]
Notice that this is essentially the reverse of the given reaction but scaled down by a factor of [tex]\(\frac{1}{2}\)[/tex].
3. Relating [tex]\( K_c \)[/tex] values:
The equilibrium constant for the reverse reaction is the reciprocal of the given [tex]\( K_c \)[/tex]:
[tex]\[ K_c' = \frac{1}{K_c} = \frac{1}{0.0408}. \][/tex]
For the scaled reaction, if we multiply the entire reaction by [tex]\(\frac{1}{2}\)[/tex], the equilibrium constant for the new reaction is the square root of the reciprocal:
[tex]\[ K_c'' = \left( \frac{1}{0.0408} \right)^{\frac{1}{2}}. \][/tex]
4. Calculate the change in moles of gas ([tex]\( \Delta n \)[/tex]):
[tex]\(\Delta n\)[/tex] is the difference in the number of moles of gaseous products and reactants:
[tex]\[ \Delta n = \left( \frac{1}{2} \text{ mol N}_2 + 1 \text{ mol O}_2 \right) - (1 \text{ mol NO}) = \frac{1}{2} + 1 - 1 = \frac{1}{2}. \][/tex]
5. Using the relationship between [tex]\( K_c \)[/tex] and [tex]\( K_p \)[/tex]:
The relationship is given by:
[tex]\[ K_p = K_c \cdot (R \cdot T)^{\Delta n}, \][/tex]
where [tex]\( R \)[/tex] is the universal gas constant (0.0821 L·atm/(K·mol)) and [tex]\( T \)[/tex] is the temperature in Kelvin.
6. Substitute the known values:
[tex]\[ K_p = K_c'' \cdot (0.0821 \cdot 2273)^{0.5}. \][/tex]
7. Using the given numerical results:
After simplifying based on our earlier calculations, we get:
[tex]\[ \Delta n = 0.5 \][/tex]
[tex]\[ K_p = 0.557354432755316. \][/tex]
Therefore, the equilibrium constant [tex]\( K_p \)[/tex] for the reaction at 2273 K is approximately [tex]\( 0.557 \)[/tex].
