# Question #0b8eb

##### 2 Answers

The equilibrium *partial pressure* of **0.22 M**.

Once again, before doing any actual calculations, take a second to try and predict what's going to happen when you mix those amounts of the gases together.

Notice that the equilibrium constant, *much smaller than 1*, which implies that this equilibrium will favor the reactant almost exclusively.

You could predict that the equilibrium partial pressure of *larger* than the initial value of **0.20 atm**, which implies that the reverse reaction will proceed until equilibrium is reached.

To confirm this, calculate the **reaction quotient**,

Since **left**, favoring the formation of even more reactant. Use an **ICE table** to help you with the calculations

**I**......0.20...............0.010......0.010

**C**....(+

**E**....0.20+2x.........0.010-x....0.010-x

The equilibrium constant will be equal to

Take the square root of both sides of the equation to get

This is equivalent to

As a result, the equilibrium partial pressure of

Rounded to two sig figs, the answer will be

Indeed, the equilibrium partial pressure of **left**, that the equilibrium partial pressures of the two products will be

The concentration =

Initial partial pressures (Atm):

At equilibrium we can assume x mol of

This means that the partial pressure of

Because the value of Kp is so small I am going to assume that the reaction virtually goes to completion right to left.

This means that the equilibrium partial pressure of

Concentration