Question

The equilibrium pressures below were observed at a certain temperature for the following reaction.
$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$
PNH₃ = 3.1 ✕ 10⁻² atm
PN₂ = 8.5 ✕ 10⁻¹ atm
PH₂ = 3.1 ✕ 10⁻³ atm
Calculate the value for the equilibrium constant $K_p$ at this temperature.

Answer 1

Answer:

37 950

Explanation:

Thinking process:

Let's consider a reversible gaseous reaction:

$aA_{(g)} + bB_{(g)} = cC_{(g)} + dD_{(g)}$

As all the reactants and products are gaseous, the reaction is a homogeneous reaction. In other words, the reactants and and products are in the same phase.

The equilibrium constant is expressed in terms of partial pressures like this:

$K_{p} = \frac{(P_{C}) ^{c}) (P_{D} )^{d} }{(P_{A}) ^{a} (P_{B}) ^{b} }$

using the formula for the reaction in the question:

Kp = (3.1 ✕ 10⁻² atm)²/ (8.5 ✕ 10⁻¹ atm) (3.1 ✕ 10⁻³ atm)³

     = 37 950