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Chapter 14: Problem 83

Consider the reaction $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \\\ & \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+\mathrm{H}^{+}(a q)+\mathrm{I}^{-}(a q) \end{aligned} $$ How could you follow the progress of the reaction by measuring the electrical conductance of the solution?

Short Answer

Expert verified
The progress of the reaction can be followed by measuring the electrical conductance of the solution. As the reaction takes place, more ions (H+ and I-) are formed which in turn increases the electrical conductivity of the solution. Therefore, by measuring the change in electrical conductance, the progress of the reaction can be followed.

Step by step solution

01

Understand the Reaction, Formation of Ions, and Electrical Conductance

Apply the knowledge of chemical reactions and understand the given equation. After the reaction takes place, the product side includes H+ (a cation) and I- (an anion) which are charged particles or ions. Electricity is conducted by the movement of free ions. Hence, tracking the electrical conductance of the solution is equivalent to tracking the concentration of these ions, as electrical conductivity of a solution is directly proportional to the concentration of ions in it.
02

Role of Electrical Conductance in Tracking the Reaction

As the reaction progresses, more C2H5I will react with water to produce more C2H5OH, H+ and I-. Thus, the concentration of the ions H+ and I- in the solution will increase. As a result, the electrical conductance of the solution will increase since water solutions conduct electricity through the movement of ions.
03

Measuring using an Electrical Conductance Meter

Using an electrical conductance meter, measure the conductivity. As the reaction proceeds and more ions are formed, the electrical conductance should increase. So, by observing the increase in electrical conductance, the progress of the reaction can be followed.

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Most popular questions from this chapter

The thermal decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) into phosphorus and molecular hydrogen is a first-order reaction: $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ The half-life of the reaction is 35.0 s at \(680^{\circ} \mathrm{C}\). Calculate (a) the first-order rates constant for the reaction and (b) the time required for 95 percent of the phosphine to decompose.

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