Question

A 1 L container originally holds 0.4 mol of N2, 0.1 mol of O2, and 0.08 mole of NO. If the volume of the container holding the equilibrium mixture of N2, O2, and NO is decreased to 0.5 L without changing the quantities of the gases present, how will their concentrations change

Answer 1

Answer:

a)  $N_2=0.8mol/L$

b)  $O_2=0.2mol/L$

c)  $NO=0.16mol/L$

Explanation:

From the question we are told that:

Moles 0f Nitrogen $N_2=0.4$

Moles 0f Oxygen $O_2=0.1$

Volume Decrease $V_2=0.5L$

Generally, the equation for Concentration is mathematically given by

$C=\frac{moles}{V}$

For Nitrogen

$N_2=\frac{0.4}{0.5}$

$N_2=0.8mol/L$

For Oxygen

$O_2=\frac{0.1}{0.5}$

$O_2=0.2mol/L$

For Nitrogen

$NO=\frac{0.08}{0.5}$

$NO=0.16mol/L$