To determine the reaction quotient (Q) for the given chemical equilibrium, we need to use the partial pressures of the reactants and products. The reaction in question is:
[tex]\[ 2 \, \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) \][/tex]
Given:
- Partial pressure of NO[tex]\(_2\)[/tex] is 0.30 bar.
- Partial pressure of N[tex]\(_2\)[/tex]O[tex]\(_4\)[/tex] is 1.0 bar.
- The equilibrium constant [tex]\(K_p\)[/tex] is 0.25.
Firstly, let's recall the formula for the reaction quotient [tex]\(Q\)[/tex]. For the general reaction:
[tex]\[ aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g) \][/tex]
The reaction quotient [tex]\(Q\)[/tex] is given by:
[tex]\[ Q = \frac{(\text{partial pressure of } C)^c \cdot (\text{partial pressure of } D)^d}{(\text{partial pressure of } A)^a \cdot (\text{partial pressure of } B)^b} \][/tex]
For our specific reaction:
[tex]\[ 2 \, \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) \][/tex]
The reaction quotient [tex]\(Q\)[/tex] becomes:
[tex]\[ Q = \frac{(\text{partial pressure of } N_2O_4)}{(\text{partial pressure of } NO_2)^2} \][/tex]
Now, substituting the given values:
Partial pressure of NO[tex]\(_2\)[/tex] = 0.30 bar
Partial pressure of N[tex]\(_2\)[/tex]O[tex]\(_4\)[/tex] = 1.0 bar
[tex]\[ Q = \frac{1.0 \, \text{bar}}{(0.30 \, \text{bar})^2} \][/tex]
[tex]\( (0.30 \, \text{bar})^2 = 0.09 \, \text{bar}^2 \)[/tex]
Thus,
[tex]\[ Q = \frac{1.0 \, \text{bar}}{0.09 \, \text{bar}^2} \][/tex]
[tex]\[ Q = 11.1111 \][/tex]
Therefore, the reaction quotient [tex]\(Q\)[/tex] for the reaction is [tex]\( 11.1111 \)[/tex].
