Answer:
Explanation:
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In this case, since the undergoing chemical reaction is:
[tex]2NO+O_2\rightleftharpoons 2NO_2[/tex]
The equilibrium expression in terms of pressures is:
[tex]Kp=\frac{p_{NO_2}^2}{p_{NO}^2p_{O_2}}[/tex]
Thus, for the initial conditions, we compute the initial pressures of both nitric oxide and oxygen:
[tex]p_{NO}^{0}=\frac{12.0g*0.082\frac{atm*L}{mol*K}*298.15K}{30g/mol*10.0L}=0.978atm\\\\ p_{O_2}^{0}=\frac{12.0g*0.082\frac{atm*L}{mol*K}*298.15K}{32g/mol*10.0L}=0.917atm[/tex]
Next, since the equilibrium pressure is 1148 mmHg or 1.51 atm, we can write:
[tex]p_T=p_{NO_2}+p_{NO}+p_{O_2}\\\\1.51=2x+0.978-2x+0.917-x\\\\1.51=1.90-x\\\\x=0.39atm[/tex]
Thus, the Kp turns out:
[tex]Kp=\frac{(2*0.39)^2}{(0.978-2*0.39)^2(0.917-x)} \\\\Kp=29.4[/tex]
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