Question

Bond dissociation enthalpy of Az(g), B2(g) and AB(g)
are x, y and z kJ mol-' respectively. The enthalpy of
formation of AB is
2x + 2y – z kJ mol-1
X + y + 2z kJ mol-1
X.V-2 kJ mol-1
2 2
2z
X + y - z kJ mol-1​

Answer 1

Answer:

X.V-2 kJ mol-1

2 2

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Answer 2

Concept:

The ordinary change in enthalpy that happens when a bond is gotten through homolytic splitting is known as bond separation enthalpy. Revolutionaries normally result from the homolysis of the bond as the finished result. How much energy held in a connection between particles in an atom is known as bond enthalpy, otherwise called bond-separation enthalpy, normal bond energy, or bond strength. It alludes explicitly to the energy expected to separate a security in the gas deliberately ease in a homolytic or balanced way.

Given:

Bond dissociation enthalpy of A2(g), B2(g) and AB(g) are $x, y and z kJ mol-'$ respectively.

Find:

We are asked to find the enthalpy of formation of AB

Solution:

According to the question,

Bond dissociation enthalpy of A2(g), B2(g) and AB(g) are $x, y and z kJ mol-'$ respectively.

So, the formation enthalpy is $X + y - z kJ mol-1$

Hence the formation enthalpy is $X + y - z kJ mol-1$.

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