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Chemistry Question 129 – AP-EAMCET 2025

The enthalpy of atomization of CH3NH2(g) is 2313 kJ mol1. If ΔCHH and ΔNHH are 414 and 389 kJ mol1 respectively, then ΔCNH (in kJ mol1) will be

The enthalpy of atomization is the total energy required to break all the bonds in one mole of a gaseous substance to form individual gaseous atoms.

Step 1: Identify Bonds and Their Counts in Methylamine✦ Active

Methylamine (CH3NH2) has the following structure: one carbon atom bonded to three hydrogen atoms and one nitrogen atom, and the nitrogen atom is bonded to two hydrogen atoms. Therefore, the molecule contains:

- Three CH bonds

- Two NH bonds

- One CN bond

💡 Teacher's Secret Hint

Visualizing the Lewis structure or structural formula of the molecule is crucial to correctly count the number of each type of bond.

Step 2: Formulate the Enthalpy of Atomization Equation○ Expand

The enthalpy of atomization (ΔatomizationH) is the sum of the average bond enthalpies of all the bonds in the molecule.

ΔatomizationH=(3×ΔCHH)+(2×ΔNHH)+(1×ΔCNH)
💡 Teacher's Secret Hint

Remember that bond enthalpies are always positive values, as energy is required to break bonds.

Step 3: Substitute Known Values○ Expand

Given:

- ΔatomizationH=2313 kJ mol1

- ΔCHH=414 kJ mol1

- ΔNHH=389 kJ mol1

Substitute these values into the equation:

2313=(3×414)+(2×389)+ΔCNH
💡 Teacher's Secret Hint

Pay close attention to the stoichiometry (number of each type of bond) when multiplying bond enthalpies.

Step 4: Solve for the C-N Bond Enthalpy○ Expand

Perform the multiplications and then solve for ΔCNH:

2313=1242+778+ΔCNH
2313=2020+ΔCNH
ΔCNH=23132020
ΔCNH=293 kJ mol1
💡 Teacher's Secret Hint

Double-check your arithmetic, especially when dealing with multiple additions and subtractions. The units for bond enthalpy are typically kJ mol1.

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