Question

a chemist is studying the following equilibrium, which has the given equilibrium constant at a certain temperature:

2no(g) + cl₂(g) ⇌ 2nocl(g) kₚ = 1.×10⁻⁶

he fills a reaction vessel at this temperature with 14. atm of nitrogen monoxide gas and 17. atm of chlorine gas. use this data to answer the questions in the table below.

can you predict the equilibrium pressure of nocl, using only the tools available to you within aleks?

if you said yes, then enter the equilibrium pressure of nocl at right. round your answer to 1 significant digit.

Explanation:

Step1: Set up the ICE table

Let \(x\) be the change in pressure for \(Cl_2\). The initial pressures are \(P_{NO}=14\ atm\), \(P_{Cl_2}=17\ atm\), and \(P_{NOCl} = 0\ atm\).
The equilibrium pressures are \(P_{NO}=(14 - 2x)\ atm\), \(P_{Cl_2}=(17 - x)\ atm\), and \(P_{NOCl}=2x\ atm\).
The equilibrium constant expression \(K_p=\frac{P_{NOCl}^2}{P_{NO}^2P_{Cl_2}}\)
Substituting the values, we get \(1\times10^{-6}=\frac{(2x)^2}{(14 - 2x)^2(17 - x)}\)
Since \(K_p\) is very small (\(1\times 10^{-6}\)), we can assume that \(x\) is very small compared to the initial pressures. So \(14-2x\approx14\) and \(17 - x\approx17\)

Step2: Simplify the equilibrium constant expression

The expression becomes \(1\times10^{-6}=\frac{4x^2}{14^2\times17}\)
First, calculate \(14^2\times17 = 196\times17=3332\)
Then, \(4x^2=1\times10^{-6}\times3332\)
\(4x^2 = 3.332\times10^{-3}\)
\(x^2=\frac{3.332\times10^{-3}}{4}=8.33\times10^{-4}\)
\(x=\sqrt{8.33\times10^{-4}}\approx0.0289\)
The equilibrium pressure of \(NOCl\) is \(2x\), so \(P_{NOCl}=2\times0.0289 = 0.0578\approx0.06\)

Answer:

yes, \(0.06\) atm