Consider the following reaction where kp = 4.55×10-5 at 723 k:

n2(g) + 3h2(g) ⇋ 2nh3(g)
a) if the three gases are mixed in a rigid container at 723 k so that the partial pressure of each gas is initially one atm, what will happen?
b) indicate true (t) or false (f) for each of the following:

1. a reaction will occur in which nh3(g) is consumed.
2. kp will increase.
3. a reaction will occur in which n2 is produced.
4. q is less than k.
5. the reaction is at equilibrium. no further reaction will occur.

True, False, True, False, False

Explanation:

The ebst way to know this is first to calculate the value of Q. This is a value that can be calculated the same way we calculate Kp:

Kp = pNH3² / pN2 * pH2³

Q = pNH3² / pN2 * pH2³

The difference between these two values is that Q is used when the reaction has not reached it's equilibrium state. We know the innitial partial pressures of all gases in the reaction, so we can calculate Q.

Q = 1² / 1 * 1³ = 1

Now, that we have this value, we need to know the meaning so we can answer the true and false question. So we need to compare with the value of Kp.

AS we can see, Q > Kp, because 1 > 4.55x10^-5, therefore, the reaction is balanced but is all oriented to the reactants and not the products, therefore, NH3 is consumed rather than being produce, so the first one is True.

For the second one, Kp is value that it's constant, the only way we can increase or decrease this value, is altering the pressure, concentration or volume of the reaction, and this is not taking place, so second is False

The third, as we state in the first one, the reaction goes to the left, rather than go to the formation of produced, therefore, N2 is produced. Third is True.

Fourth, we already do the calculations and Q > Kp, so fourth is False

Finally, indeed the reaction is in equilibrium but according to the value of Q and previous explanation, we can see that there is still reaction in this equilibrium, and fifth is False

particle   p is an electron and particle q is a neutron