Calculate the concentration of \(\mathrm{CO}_{2}\) in a soft drink that is bottled with a partial pressure of \(\mathrm{CO}_{2}\) of \(4.0 \mathrm{~atm}\) over the liquid at \(25^{\circ} \mathrm{C}\). The Henry's law constant for \(\mathrm{CO}_{2}\) in water is \(3.1 \times 10^{-2} \mathrm{~mol} /\) litre-atm at this temperature.

Short Answer

Expert verified
The concentration of \( \mathrm{CO}_{2} \) in the soft drink is \( 0.124 \mathrm{~mol} / \text{liter} \).

Step by step solution

01

Understand Henry's Law

Henry's law states that the solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of that gas above the liquid. The law is expressed as: \( C = kP \), where \( C \) is the concentration of the gas in the liquid, \( k \) is Henry's law constant, and \( P \) is the partial pressure of the gas.
02

Identify Given Values

The partial pressure of \( \mathrm{CO}_{2} \) (\( P \)) is given as \( 4.0 \mathrm{~atm} \) and Henry's law constant (\( k \)) for \( \mathrm{CO}_{2} \) in water is \( 3.1 \times 10^{-2} \mathrm{~mol} / \text{litre-atm} \) at \( 25^\circ \mathrm{C} \).
03

Apply Henry's Law

Now apply Henry's law formula \( C = kP \) with the given values. Substitute \( k = 3.1 \times 10^{-2} \mathrm{~mol} / \text{litre-atm} \) and \( P = 4.0 \mathrm{~atm} \) into the equation to find the concentration \( C \) of \( \mathrm{CO}_{2} \).
04

Calculate the Concentration of CO2

The concentration of \( \mathrm{CO}_{2} \) is calculated as follows: \[ C = (3.1 \times 10^{-2} \mathrm{~mol} / \text{litre-atm}) \times (4.0 \mathrm{~atm}) \]
05

Solve for Concentration C

Multiplying the values, we have: \[ C = (3.1 \times 10^{-2} \times 4.0) \mathrm{~mol} / \text{litre} = 12.4 \times 10^{-2} \mathrm{~mol} / \text{litre} \] Simplifying, we get the concentration of \( \mathrm{CO}_{2} \) in the soft drink as \( 0.124 \mathrm{~mol} / \text{litre} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gas Solubility
Understanding the solubility of gases in liquids is a fundamental aspect of physical chemistry, relevant to many practical applications such as carbonation of beverages.

Solubility is the ability of a gas to dissolve in a liquid under certain conditions. In general, solubility increases with an increase in pressure and decreases with an increase in temperature. However, each gas-liquid combination has its unique characteristics.

When we talk about gas solubility in water, for instance, Henry's Law becomes a critical point of reference. It specifically relates the amount of gas that dissolves in a liquid with the partial pressure of the gas above the liquid; this allows the calculation of dissolved gas concentrations using the constant provided for the particular gas at a specific temperature.
Partial Pressure
Partial pressure is another essential concept in the study of gas solubility. It refers to the pressure exerted by a single gas in a mixture of gases. For example, in a bottle of carbonated drink, not all the pressure is due to carbon dioxide; there are other gases present which contribute to the total pressure.

In Henry's Law, we are interested in the partial pressure of the specific gas that is dissolving in the liquid. The law clearly demonstrates that a higher partial pressure of the gas results in greater solubility in the liquid. This relation is of paramount importance while considering gas-liquid interactions and is particularly essential in the manufacturing of soft drinks, as illustrated in the exercise.
Concentration Calculation
The concentration of a gas in a liquid is quantifiable, and determining this value is crucial in fields like environmental science, food chemistry, and industrial processes. Concentration is typically expressed as mole per liter (mol/L), which represents how much of a substance is present in a given volume of liquid.

The step-by-step solution provides an excellent example of how to apply Henry's Law for concentration calculation. By multiplying the Henry's Law constant by the partial pressure of the gas, we arrive at the concentration of the gas dissolved in the liquid. In this case, the calculation for \( \text{CO}_2 \) concentration in the soft drink leads to a straightforward multiplication, demonstrating the direct proportionality outlined by Henry's Law.
Physical Chemistry for IIT-JEE
For students preparing for the Indian Institutes of Technology Joint Entrance Examination (IIT-JEE), grasping the principles of physical chemistry is pivotal. This challenging exam tests the deep understanding of concepts like gas solubility and its relation to partial pressure, as seen with Henry's Law.

Exercises like the one illustrated help students to apply theoretical knowledge to practical scenarios, promoting critical thinking and problem-solving skills. It's not merely about plugging values into an equation; it's about comprehending the underpinning physical principles that govern these reactions. Doing so is valuable for acing competitive exams such as the IIT-JEE, which demands mastery over the application of these scientific principles.

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Most popular questions from this chapter

Aluminium crystallizes in a cubic close-packed structure. Its metallic radius is \(125 \mathrm{pm}\). (a) What is the length of the side of the unit cell? (b) How many unit cells are there in \(1.00 \mathrm{~cm}^{3}\) of aluminium?

Sodium metal crystallizes in a body centred cubic lattice with the cell edge \(a=4.29 \AA\). What is the radius of sodium atom?

Which of the following reactions will get affected by increase of pressure? Also mention, whether change will cause the reaction to go into the right or left direction? (i) \(\mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{~S}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CS}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{~S}_{(\mathrm{g})}\) (ii) \(\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{C}_{(\mathrm{s})} \rightleftharpoons 2 \mathrm{CO}_{(\mathrm{g})}\) (iii) \(4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\) (iv) \(\mathrm{C}_{2} \mathrm{H}_{4(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6(\mathrm{~g})}\)

A compound formed by elements \(A\) and \(B\) crystallizes in cubic structure where \(A\) atoms are at the corners of a cube and \(B\) atoms are at the face centre. What is the formula of the compound?

\(\Delta G^{\circ}\) for \(\frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2} \rightleftharpoons \mathrm{NH}_{3}\) is \(-16.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(25^{\circ} \mathrm{C}\). Find out \(K_{\mathrm{p}}\) for the reaction. Also report \(K_{\mathrm{p}}\) and \(\Delta G^{\circ}\) for: $$ \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \quad \text { at } 25^{\circ} \mathrm{C} $$

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