Question

For the reaction of nitric oxide and oxygen to form nitrogen dioxide, the reaction begins with 12.0 g of nitric oxide and 12.0 g of oxygen at 25oC in a 10.0 L container. At equilibrium, the pressure in the container is 1148 mmHg, what is Kp?

Answer 1

Step-by-step explanation:

Hello!

In this case, since the undergoing chemical reaction is:

`2NO+O_2\rightleftharpoons 2NO_2`

The equilibrium expression in terms of pressures is:

`Kp=(p_(NO_2)^2)/(p_(NO)^2p_(O_2))`

Thus, for the initial conditions, we compute the initial pressures of both nitric oxide and oxygen:

`p_(NO)^(0)=(12.0g*0.082(atm*L)/(mol*K)*298.15K)/(30g/mol*10.0L)=0.978atm\\\\ p_(O_2)^(0)=(12.0g*0.082(atm*L)/(mol*K)*298.15K)/(32g/mol*10.0L)=0.917atm`

Next, since the equilibrium pressure is 1148 mmHg or 1.51 atm, we can write:

`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`

Thus, the Kp turns out:

`Kp=((2*0.39)^2)/((0.978-2*0.39)^2(0.917-x)) \\\\Kp=29.4`

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