Chapter 5, Problem 5.61PAE

61 As one step in its purification, nickel metal reacts with carbon monoxide to form a compound called nickel tetracarbonyl, Ni(CO)₄, which is a gas at temperature above about 316 K. A 2.000-L flask is filled with CO gas to a pressure of 748 torr at 350.0 K, and then 5.00 g of Ni is added. If the reaction describe occurs and goes to completion at constant temperature, what will the final pressure in the falsk be?

Interpretation Introduction

To determine:

$Ni+4CO\to Ni{\left(CO\right)}_4$

A $2.00L$ flask is filled with $CO$ gas to a pressure of $748\text{ torr}$ at $350.0k,\text{ then 5}\text{.00g }Ni$ is added. If this reaction proceeds at constant temperature, then find the final pressure in the flask.

Explanation of Solution

$\begin{array}{c}PV=nRT\\ n=\frac{PV}{RT}\end{array}$

$\begin{array}{c}P=748\text{ torr}\\ =\frac{748}{760}atm\left(760\text{ torr}=\text{1atm}\right)\\ =\frac{187}{190}atm\\ T=350\text{ K}\\ \text{V}=\text{2}\text{.00 L}\\ =2.00\times {10}^{-3}{m}^3\\ R=8.314{\text{ Jmol}}^{-1}{K}^{-1}\\ n=\frac{\left(\frac{187}{190}\right)atm\times \left(2.00\times {10}^{-3}{m}^3\right)}{\left(8.314{\text{ Jmol}}^{-1}{K}^{-1}\right)\times \left(350\text{ K}\right)}\\ =6.67\times {10}^{-7}mol\end{array}$

$\begin{array}{c}\text{moles of }Ni=\frac{\text{mass of Ni}}{\text{molar mass of }Ni}\\ =\frac{5g}{58.7g/m}\\ =8.5\times {10}^{-2}mol.\end{array}$

From the above calculation it becomes clear that limit reagent is $CO$ i.e. whole $CO$ gas will convert into $Ni{\left(CO\right)}_4.$.

$Ni\left(S\right)+4CO\left(g\right)\to Ni{\left(CO\right)}_4\left(g\right)$

$4$ moles of $CO$ will produce $1$ mole of $Ni{\left(CO\right)}_4\left(g\right)$.

We cause it given that temperature remain cost

$P_i{V}_i=n_{i}R{T}_i\cdots \cdots \left(1\right)$

$i$ : initial parameter before reaction

$P_f{V}_f=n_{f}R{T}_f\cdots \cdots \left(1\right)$

$f$ : final parameter after reaction

Let initial mole of $CO=ni$

Final moles of $Ni{\left(CO\right)}_4=\frac{ni}{4}$.

$\begin{array}{c}{T}_i=T_f\left(\text{given}\right)\\ V_i=V_f\left(\text{volume remain same}\right)\\ \raisebox{1ex}{$\left(1\right)$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ \frac{P_i{V}_i}{P_f{V}_f}=\frac{n_{i}R{T}_i}{n_{f}R{T}_f}\\ \frac{P_i}{P_f}=\frac{n_i}{\frac{n_i}{4}}=4\\ P_f=\frac{P_i}{4}\\ \left(P_i=748\text{ torr}\right)\\ =\frac{748\text{ torr}}{4}\\ =187\text{ torr}\end{array}$

Hence final pressure $=187\text{ torr}$

Conclusion

At fixed temperature and volume pressure is directly proportional to no of moles of gas.