Answered By Leonardo F

Question 1) The first thing we need to do is find out how many moles of NO actually reacted, given that we placed in the flask 2.5 mol of NO. We can do this by using the following equation for reagents:

Chemical Equilibrium amount = initial amount - amount reacted

For products:

Chemical Equilibrium amount = initial amount + amount formed

In the case of  $N_2O$N₂O , in the chemical equilibrium, we have 0.58 mol of it. This means that, since we don't have an initial amount of  $N_2O$N₂O , the amount formed is also 0.58 mol. We can now calculate the amount of moles reacted of NO, given that their molar proportion is 4 mol of NO for 2 mol of  $N_2O$N₂O :

 $\frac{4}{x}=\frac{2}{0.58}\rightarrow x=\frac{4\times0.58}{2}=1.16$4x =20.58 →x=4×0.582 =1.16 

So, the amount of NO reacted is 1.16 mol. We can also calculate the amount formed of oxygen (molar proportion of 4 mol of NO: 1 mol of oxygen):

 $\frac{4}{1.16}=\frac{1}{y}\rightarrow y=\frac{1.16\times1}{4}=0.29$41.16 =1y →y=1.16×14 =0.29 

So, the amount formed of oxygen is 0.29 mol. Let's now calculate the amount of moles of each component in the chemical equilibrium:

  $NO\rightarrow2.5-1.16=1.34$NO→2.5−1.16=1.34 mol

 $N_2O\rightarrow0+0.58=0.58$N₂O→0+0.58=0.58 mol

 $O_2\rightarrow0+0.29=0.29$O₂→0+0.29=‍0.29 mol

The percent yield is calculated by:

Yield(%)=(actual yield/theoretical yield)*100

The actual yield we already know: 0.58 mol of  $N_2O$N₂O . The theoretical yield can be calculated using the molar proportion of 4:2 between NO and  $N_2O$N₂O . Since we initially put on the flask 2.5 mol of NO:

 $\frac{4}{2.5}=\frac{2}{z}\rightarrow z=\frac{2.5\times2}{4}=1.25$42.5 =2z →z=2.5×24 =1.25 mol

This means that if all the 2.5 mol of NO reacted, we would have formed 1.25 mol of  $N_2O$N₂O (theoretical yield). Hence:

 $Yield=\frac{0.58}{1.25}=0.46=46\%$Yield=0.581.25 =0.46=46% 

Since the total volume in the flask is 2.0 L, if we divide each amount above by 2.0 L, we will have the equilibrium concentrations (in mol/L) of each component:

For NO:  $\frac{1.34mol}{2.0L}=0.67$1.34mol2.0L =0.67 mol/L

For N2O:  $\frac{0.58mol}{2.0L}=0.29$0.58mol2.0L =0.29 mol/L

For O2:   $\frac{0.29mol}{2.0L}=0.15$0.29mol2.0L =0.15  mol/L

Hence, the equilibrium concentration of NO is 0.67 mol/L.

The equilibrium constant (kc) is given by:

 $kc=\frac{\left[N_2O\right]^2\left[O_2\right]}{\left[NO\right]^4}$kc=[N₂O]²[O₂][NO]⁴  

Putting the values:

 $kc=\frac{\left(0.29\right)^2\left(0.15\right)}{\left(0.67\right)^4}=0.061$kc=(0.29)²(0.15)(0.67)⁴ =0.061 

In scientific notation:

 $kc=6.1\times10^{-2}$kc=6.1×10⁻² 

Question 2) If we add to the mixture barium nitrate, in water, it dissociates:

 $Ba\left(NO_3\right)_2\rightarrow Ba^{2+}+2NO_3^-$Ba(NO₃)₂→Ba²⁺+2NO₃⁻ 

These ions shown in the chemical equation above can combine with copper and sulfate ions, forming copper (II) nitrate ($Cu\left(NO_3\right)_2$Cu(NO₃)₂) and barium sulfate ($BaSO_4$‍BaSO₄). Copper (II) nitrate is soluble in water, but barium sulfate is not. It precipitates in the solid form, which means that if we add barium nitrate to the original mixture, we will be decreasing the concentration of sulfate ($SO_4^{2-}$SO₄²⁻) ions, and the equilibrium will be shifted to the right, because when we decrease the concentration of a participant in the reaction, the equilibrium is shifted to the same side of this participant. Since the equilibrium is shifted to the right, we will be decreasing the concentration of solid copper(II) sulfate (since it's on the left side of the equation).

Hence, the blanks are filled with "decrease" and "right".