Equilibrium NEET 2007: Combining Equilibrium Constants

Equilibrium NEET 2007: what constant are we being asked to find?

The balanced target reaction fixes every exponent in the answer. In this Equilibrium NEET 2007 question, combine three supplied equilibria to find the constant for converting ammonia and oxygen into nitric oxide and water:

2 N H_{3} + \frac{5}{2} O_{2} ⇌ 2 N O + 3 H_{2} O

This 2007 Chemistry Equilibrium MCQ is tagged hard on this question bank’s scale, not on the basis of student response data. Use 90 seconds as a practice benchmark, not as a measured average solving time.

The supplied reactions and their constants are:

N_{2} + 3 H_{2} ⇌ 2 N H_{3} K_{1}

N_{2} + O_{2} ⇌ 2 N O K_{2}

H_{2} + \frac{1}{2} O_{2} ⇌ H_{2} O K_{3}

Choose an option before reading the solution. Write your reaction operations beside your choice so you can check the reasoning later.

Option A

\frac{K_{2} K_{3}^{3}}{K_{1}}

Option B

\frac{K_{2} K_{3}^{2}}{K_{1}}

Option C

\frac{K_{2}^{2} K_{3}}{K_{1}}

Option D

\frac{K_{1} K_{2}}{K_{3}}

How do we match each supplied reaction to the target?

2 N H_{3} + \frac{5}{2} O_{2} ⇌ 2 N O + 3 H_{2} O

Reverse the first reaction, retain the second and triple the third. These operations put ammonia on the reactant side and produce the required nitric oxide and water. The combination rules apply to equilibrium constants at the same temperature.

Two ammonia molecules supply two nitrogen atoms, requiring two nitric oxide molecules. They also supply six hydrogen atoms, requiring three water molecules: $Hydrogen atoms = 2 \times 3 = 6$

Water coefficient = \frac{6}{2} = 3

The products contain five oxygen atoms. This fixes the oxygen coefficient:

Oxygen atoms = 2 \times 1 + 3 \times 1 = 5

$Oxygen molecule coefficient = \frac{5}{2}$ Use this ledger to pair each reaction operation with its modified equation and modified constant:

Reverse reaction 1. Reversing exchanges reactants and products, so the constant becomes its reciprocal.

2 N H_{3} ⇌ N_{2} + 3 H_{2} \frac{1}{K_{1}}

Keep reaction 2 unchanged. Its direction already produces nitric oxide, so its constant stays unchanged.

N_{2} + O_{2} ⇌ 2 N O K_{2}

Multiply reaction 3 by three. Scaling every coefficient raises the constant to that power.

3 H_{2} + \frac{3}{2} O_{2} ⇌ 3 H_{2} O K_{3}^{3}

The factor of three comes from cancellation, not the options. Reversed ammonia formation releases three hydrogen molecules, and water formation must consume all three. Tripling gives:

K_{3}^{3}, not 3 K_{3}

How do cancellation and multiplication give option A?

The nitrogen and hydrogen intermediates cancel, leaving exactly the target reaction. Adding equilibrium reactions then multiplies their modified constants; it does not add the constants.

The unsimplified sum is:

& 2 N H_{3} + N_{2} + O_{2} + 3 H_{2} + \frac{3}{2} O_{2} & ⇌ N_{2} + 3 H_{2} + 2 N O + 3 H_{2} O

Cancel nitrogen from opposite sides, then cancel the three hydrogen molecules from opposite sides. Neither intermediate remains in the target.

The remaining oxygen coefficients add:

1 + \frac{3}{2} = \frac{2}{2} + \frac{3}{2} = \frac{5}{2}

This recovers exactly:

2 N H_{3} + \frac{5}{2} O_{2} ⇌ 2 N O + 3 H_{2} O

Multiply the modified constants:

K = \frac{1}{K_{1}} \times K_{2} \times K_{3}^{3} = \frac{K_{2} K_{3}^{3}}{K_{1}}

Option A is correct. The denominator comes from reversing reaction 1; the cube comes from tripling reaction 3.

An equilibrium constant belongs to the equation as written. Doubling the whole target to remove the fraction gives:

4 N H_{3} + 5 O_{2} ⇌ 4 N O + 6 H_{2} O

That equation has the square of the original constant, not the same constant. Removing a fractional coefficient is not just a change in presentation: scaling the equation changes its equilibrium constant.

Why does option B give the wrong equilibrium constant?

Option B uses a square where the target requires a cube. One way to generate it is to write the tripled water-formation equation but attach the constant for doubling it.

Option B is:

\frac{K_{2} K_{3}^{2}}{K_{1}}

The mismatched equation-and-constant pair is:

3 H_{2} + \frac{3}{2} O_{2} ⇌ 3 H_{2} O \underset{incorrect: must be K_{3}^{3}}{\underset{⏟}{K_{3}^{2}}}

Alternatively, interpret the square consistently: use water formation only twice. That consumes two hydrogen molecules, leaving one hydrogen molecule uncancelled. The actual net reaction becomes:

2 N H_{3} + 2 O_{2} ⇌ 2 N O + 2 H_{2} O + H_{2}

This is a different reaction, so its constant cannot answer the question. The square is valid for that combination, not for the requested target.

Before choosing an option, confirm that both nitrogen and hydrogen intermediates disappear and three water molecules remain. Check the net equation rather than choosing an exponent from memory.

How can I apply the same method to three related questions?

Write the reaction operations, check cancellation, then calculate the constant. These are original related practice questions, not additional verified PYQs. Attempt each before reading its worked answer.

What is the constant for the reverse oxidation reaction?

2 N O + 3 H_{2} O ⇌ 2 N H_{3} + \frac{5}{2} O_{2}

Take the reciprocal of the solved target’s constant. Every reactant and product has exchanged sides, but no coefficient has changed. There is no extra power to apply.

K_{r e v e r s e} = \frac{1}{K} = \frac{K_{1}}{K_{2} K_{3}^{3}}

What is the constant when every target coefficient is doubled?

4 N H_{3} + 5 O_{2} ⇌ 4 N O + 6 H_{2} O

Square the entire original constant. Every target coefficient is doubled. The power must act on the denominator as well as the numerator.

K_{d o u b l e} = {(\frac{K_{2} K_{3}^{3}}{K_{1}})}^{2} = \frac{K_{2}^{2} K_{3}^{6}}{K_{1}^{2}}

What is the constant for converting nitric oxide and hydrogen into nitrogen and water?

2 N O + 2 H_{2} ⇌ N_{2} + 2 H_{2} O

Reverse reaction 2 and double reaction 3. Ammonia is absent from this target, so reaction 1 is unnecessary. Use the original supplied reactions as follows:

Reverse reaction 2.

2 N O ⇌ N_{2} + O_{2} \frac{1}{K_{2}}

Double reaction 3.

2 H_{2} + O_{2} ⇌ 2 H_{2} O K_{3}^{2}

Add the equations:

2 N O + 2 H_{2} + O_{2} ⇌ N_{2} + O_{2} + 2 H_{2} O

Cancel oxygen from opposite sides to recover the requested target. Multiply the modified constants:

K_{n e w} = \frac{1}{K_{2}} \times K_{3}^{2} = \frac{K_{3}^{2}}{K_{2}}

For your next attempt, write the operation beside each reaction before touching the options: reverse means reciprocal, scale means power, add means multiply.

Next step: photograph a doubt on NEET JEEnius AI and photograph any question you are stuck on and get a step-by-step solution across Physics, Chemistry and Biology (20 free a month).

For a worked example of the same idea, see Electromagnetic Induction NEET 2026: Solenoid Inductance.

Frequently asked questions

What is the correct answer to the Equilibrium NEET 2007 question?

Option A is correct: K = K2K3^3/K1 for 2NH3 + (5/2)O2 ⇌ 2NO + 3H2O. Reverse the ammonia-formation reaction, keep the nitric-oxide-formation reaction unchanged and triple the water-formation reaction.

Why is K3 cubed instead of squared?

The target reaction requires three water molecules, so the supplied water-formation reaction must be tripled. Tripling all reaction coefficients raises its equilibrium constant to the third power, giving K3^3. Using that reaction only twice leaves one hydrogen molecule uncancelled and produces a different target reaction.

What are the rules for combining equilibrium constants?

Reversing a reaction replaces its equilibrium constant with its reciprocal. Multiplying every coefficient by a factor raises the constant to that power. Adding reactions multiplies their modified constants, provided all constants refer to the same temperature.

Does doubling a reaction change its equilibrium constant?

Doubling every coefficient squares the entire equilibrium constant. For 4NH3 + 5O2 ⇌ 4NO + 6H2O, the constant is K2^2K3^6/K1^2. Removing the fractional oxygen coefficient by doubling therefore changes the constant; it is not merely a presentation change.