Law of Equivalence: Definition, Formula, Questions and Examples

Thus, inventors resort to patent systems for the protection of their works against unauthorized usage by contemporaries using the fast-evolving landscape of innovation and technological development. But in so doing, the very nature of the evolution of the technology has made it very difficult—indeed well-nigh impossible—to ascertain with full precision the exact scope of the patent claim. Such is the place of the Doctrine of Equivalents—a legal device to protect a patentee, even in the face of additional ways to infringe, either by device or process, that are not within the literally expressed scope of the patent claim.

This Story also Contains

1. Doctrine of Equivalents
2. Types of Equivalents
3. Application in real life and its implication
4. Some Solved Examples
5. Conclusion

The Doctrine of Equivalents is designed to prevent courts from finding a party not infringing when the accused product or process is only insubstantially different from the patented invention, not allowing an infringer to steal the benefits of the invention by concocting minor changes to avoid literal infringement. The discussion that follows will, therefore, make the points of the Doctrine of Equivalents very crystal clear regarding basic principles, many dimensions, and some practical real-world applications in both the business and academic worlds.

Doctrine of Equivalents

The core idea behind the Equivalents Doctrine is that a patent cannot be restricted to the literal terms or parameters but extends further to include equivalents of the claimed invention. It has been interpreted to mean that even a device or process otherwise performing a function the same way and achieving nearly the same result as the patented invention shall be an equivalent even though given a different name or is differently shaped or formed. The doctrine is designed to withhold a person who violates the claims of the patent from making "a copy of the benefit of the invention" by small and immaterial variations.

Laws of Equivalence

According to the law of equivalence, for each and every reactant and product,
Equivalents of each reactant reacted = Equivalents of each product formed.

Example,
Suppose the reaction is taking place as follows:
$P+Q \rightarrow R+S$

According to the law of equivalence,

Equivalents of P reacted = Equivalents of Q reacted = Equivalents of R produced = Equivalents of S produced

Equivalents of any substance = (Weight of substance (in g)) / (Equivalent weight)
= Normality (N) x Volume (V) (In liter)
Normality (N) = n-Factor x Molarity (M)

Law of Equivalence finds great importance in Acid-base Neutralisation Reactions as well as Redox Titrations.

Here we shall be mainly covering the Acid-Base neutralization reactions in detail.

Types of Equivalents

The Doctrine of Equivalents really reflects two principal types of equivalents: (1) Insubstantial Differences and (2) Known Interchangeability.
Insubstantial differences are those changes that are so minor or inconsequential that they would not make a material difference in how the invention works. Known interchangeability refers to those cases where persons skilled in the art would have known that one element could have been replaced by another without affecting how the invention works.

Application in real life and its implication

The Doctrine of Equivalents has a vital application in the practical world.
It provides protection to the patentee against those infringing entities that otherwise would not infringe per se the patent in question but would remain substantially similar to the invention. For instance, if a certain company invents a new drug, the competitor cannot make some minor modifications to the formula associated with that drug and sell it under a different patent, under the Doctrine of Equivalents. The Doctrine of Equivalents is a point of rigorous study and discussion within the parameters of academia.
By using those case studies, the students will appreciate the delicate balance to be struck between providing adequate protection to the patentees and clearly notifying the public of the scope of the patent. It will also discuss the developed tests and legal bars in a bid to limit the application of the doctrine, such as the Festo Presumption and the All Elements Rule.

Recommended topic video on(Law of Equivalence)

Some Solved Examples

Example 1:In this equation:

$2 \mathrm{KClO}_3 \rightarrow 2 \mathrm{KCl}+3 \mathrm{O}_2$
What will be the mass (in g) of O₂ produced from 1.23g of KClO₃?

1) (correct)0.482

2)0.545

3)0.758

4)0.345

Solution

According to the equation,

2 moles of KClO₃ will produce 3 moles of O₂.

Thus moles of KClO₃ = 1.23 / 123= 0.01 moles

Therefore, 0.01 moles of KClO₃ will produce = (3 x 0.01) / 2= 0.015 moles

thus mass of O₂ = 0.015 x 32
= 0.48g

Hence, the answer is the option (1).

Example 2:
50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. What is the mass (in g) of NaOH in 50 mL of the given sodium hydroxide solution?

1) 20

2)4(Correct)

3)10

4)40

Solution

$\begin{aligned} & \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4+2 \mathrm{H}_2 \mathrm{O} \\ & \text { meq of } \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4=\text { meq } \mathrm{NaOH} \\ & 50 \times 0.5 \times 2=25 \mathrm{M} \mathrm{NaOH} \times 1 \\ & 1000 \mathrm{ml} \text { solution }=2 \times 240 \mathrm{gram} \mathrm{NaOH} \\ & \therefore 50 \mathrm{ml} \text { sol }=4 \mathrm{~g} \mathrm{NaOH}\end{aligned}$

Hence, the answer is the option (2).

Example 3:
The volume of 0.1 N dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04 mol of OH− in aqueous solution is:

1) 200mL

2)400mL(Correct)

3)600mL

4)800mL

Solution

Dibasic acids mean acids have two ionizable hydrogens.

gram eq. of OH^- = mole $\times$ n-factor = 0.04 $\times$ 1 = 0.04

The Law of equivalence we get,
milli equivalents of acid = milli equivalents of base

N₁ X V₁ (ACID) = ( meq. of base)

0.1×V = ( meq. of base)

0.1 X V = ( meq. of base) = 0.04 $\times$ 1000

V x 0.1 = 40
V = 400ml

Hence, the answer is an option (2).

Example 4:
How much NaOH is required to neutralize 1500 mL of 0.1 M HCl?

1) 40g

2)4g

3)6g(correct)

4)60g

Solution

According to the law of Equivalence,

meq. of HCl = meq. of NaOH

$\therefore$ 1500 $\times$ 0.1 = moles of NaOH (as n-factor of NaOH =1)

$\therefore$ moles of NaOH = 150

$\therefore$ Weight of NaOH = 6 g

Hence, the answer is the option (3).

Example 5:
To neutralize 20 mL of M10 sodium hydroxide, the volume of M20 hydrochloric acid required is:

1) 10ml

2)15ml

3)20ml

4)40ml(correct)

Solution

According to the law of equivalence,

meq. of HCl = meq. of NaOH

$\begin{aligned} & \therefore 20 \times \frac{1}{10}=\mathrm{V} \times \frac{1}{20} \\ & \mathrm{~V}=40 \mathrm{ml}\end{aligned}$

Hence, the answer is the option (4).

Conclusion

The Doctrine of Equivalents becomes an astonishing aspect of patent law in serving to protect against infringement, which will assure that the system of patents continues to encourage and reward innovation. Exponential technological progress, with all its consequent changes, shall have an impact upon us all, so will the Doctrine of Equivalents be an essential tool in the arsenal of patent holders who are looking to protect their invention from infringement but at the same time ensure there is a fair and competitive marketplace.