Question

An analysis of a Solvay-process plant shows that for every \(1.00 \mathrm{kg}\) of \(\mathrm{NaCl}\) consumed, \(1.03 \mathrm{kg}\) of \(\mathrm{NaHCO}_{3}\) are obtained. The quantity of \(\mathrm{NH}_{3}\) consumed in the overall process is \(1.5 \mathrm{kg}.\) (a) What is the percent efficiency of this process for converting NaCl to \(\mathrm{NaHCO}_{3} ?\) (b) Why is so little \(\mathrm{NH}_{3}\) required?

Short answer

(a) The percent efficiency of the process for converting NaCl to NaHCO3 is 103%. (b) Only a small amount of NH3 is required because it is a catalyst in the Solvay process and is not consumed in the reaction.

Step-by-step solution

Calculate the theoretical yield

To calculate the efficiency, the actual yield needs to be compared to the theoretical yield. Considering that NaCl and NaHCO3 have nearly equal molar masses, the theoretical yield would be equal to the amount of NaCl consumed, which is 1.00kg in this case.

Calculate the actual yield

The actual yield is the amount of NaHCO3 obtained, which is given as 1.03kg.

Calculate the percent efficiency

The percent efficiency is given by the formula: \[ Efficiency (\%) = \left( \frac{{Actual \,yield}}{{Theoretical \,yield}} \right) \times 100 \] Plugging in the values, we get \[ Efficiency (\%) = \left( \frac{{1.03}}{{1.00}} \right) \times 100 = 103\% \]

Understanding NH3 consumption

Ammonia (NH3) is a catalyst in the Solvay process and is not consumed in the reaction. That's why only a little amount of NH3 is required.

Key concepts

Percent Efficiency

In the context of chemical processes like the Solvay process, **percent efficiency** is a measure of how effectively the input materials are converted into the desired product. To calculate percent efficiency, we compare the actual yield (what we get from the process) to the theoretical yield (what we expect to get if everything works perfectly). The formula looks like this:\[Efficiency (\%) = \left( \frac{{\text{Actual yield}}}{{\text{Theoretical yield}}} \right) \times 100\]In our example, the Solvay process yields 1.03 kg of \(\text{NaHCO}_3\) from 1.00 kg of \(\text{NaCl}\). The theoretical yield is also 1.00 kg (since NaCl and NaHCO₃ have similar molar masses). By substituting these values into the efficiency formula, we find:\[Efficiency (\%) = \left( \frac{{1.03}}{{1.00}} \right) \times 100 = 103\%\]A percent efficiency of 103% indicates that the process yields more \(\text{NaHCO}_3\) than expected. This can result from impurities adding mass or calculation errors, but also highlights the effectiveness of the process in converting reactants.

Theoretical Yield

Theorizing about how much product a chemical process should produce helps us assess the process's potential efficiency. This is what **theoretical yield** represents. It's defined as the maximum amount of product expected from a reaction based on stoichiometric calculations.In the Solvay process, where \(\text{NaCl}\) is converted to \(\text{NaHCO}_3\), we assume no losses and perfect conditions. Here, the theoretical yield equals the mass of \(\text{NaCl}\) utilized, 1.00 kg. This equivalence arises because the molar masses of \(\text{NaCl}\) and \(\text{NaHCO}_3\) are nearly identical, simplifying calculations.Understanding the theoretical yield provides a baseline for measuring actual results against expectations. Deviations between actual and theoretical yield can help identify inefficiencies or areas for process improvement, serving as a vital tool in industrial chemistry.

NH3 Consumption

Ammonia (H3) plays a unique role in the Solvay process. Unlike other reactants, **NH3 consumption** is minimal, and for an important reason. NH3 is part of the reaction but works mainly as a catalyst. Catalysts are substances that increase the rate of a chemical reaction without being consumed by it. In the Solvay process, NH3 facilitates the formation of \(\text{NaHCO}_3\) and then regenerates, meaning it is reused repeatedly within the cycle. This explains why only 1.5 kg of NH3 is needed compared to larger amounts of other reagents, making the process more economical and resource-efficient.Knowing the role of NH3 helps in understanding how the Solvay process maintains efficiency while conserving catalysts, making industrial-scale production both effective and economically viable.