What is the change in molar kinetic energy for the dissociation of the diatomic gas $AB (g)$ $"AB"(g)$ into $A (g)$ $"A"(g)$ and $B (g)$ $"B"(g)$ in the high temperature limit?

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What is the change in molar kinetic energy for the dissociation of the diatomic gas $AB (g)$ $"AB"(g)$ into $A (g)$ $"A"(g)$ and $B (g)$ $"B"(g)$ in the high temperature limit?

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Answer 1 · 2017-04-15T23:51:14

Consider a general diatomic gas, $A B (g)$ $AB(g)$ , in the following dissociation reaction:

$A B (g) \to A (g) + B (g)$ $AB(g) -> A(g) + B(g)$

We say that the diatomic gas has certain degrees of freedom (DOFs):

$1$ $1$ per dimension of translational motion: $x, y, z \Rightarrow$ $x,y,z =>$ $3$ $3$ DOFs

$1$ $1$ per rotational angle: $θ$ $theta$ and $ϕ$ $phi$ $\Rightarrow$ $=>$ $2$ $2$ DOFs

$1$ $1$ per vibrational motion (there is only one: symmetrical stretch). $\Rightarrow$ $=>$ $1$ $1$ DOF

By the equipartition theorem for average kinetic energy, we have:

$K_{a v g} = \frac{N}{2} n R T$ $K_(avg) = N/2nRT$ ,

where $N$ $N$ is the degrees of freedom, $n$ $n$ is $mol$ $"mol"$ s, and $R$ $R$ and $T$ $T$ are as usual from the ideal gas law.

In total, for $A B$ $AB$ , it has six DOFs. Hence, $A B (g)$ $AB(g)$ has $K_{a v g} = \frac{1}{2} n R T$ $K_(avg) = 1/2nRT$ for each DOF, totalling $K_{a v g} = 3 n R T$ $K_(avg) = 3nRT$ .

For monatomic gases, they only have translational DOFs, so their kinetic energies are assumed to be $\frac{3}{2} n R T$ $3/2nRT$ .

When the temperature does not change, we can write the change in average kinetic energy as:

$DeltaK_(avg) = K_(avg,AB) - [K_(avg,A) + K_(avg,B)]$

$= K_(avg,AB) - 2K_(avg,A)$

$= 3nRT - 2*3/2nRT$

Therefore:

$color(blue)((DeltaK_(avg))/n) = (3-3)RT = color(blue)("0 J/mol")$

This is saying that all the energy that was partitioned into the translational, rotational, and vibrational motions of $AB(g)$ were properly conserved and transferred into the individual atomic gases $A(g)$ and $B(g)$ as the diatomic molecular gas $AB(g)$ dissociated.

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What is the change in molar kinetic energy for the dissociation of the diatomic gas $AB (g)$ $"AB"(g)$ into $A (g)$ $"A"(g)$ and $B (g)$ $"B"(g)$ in the high temperature limit?

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What is the change in molar kinetic energy for the dissociation of the diatomic gas $AB (g)$ $"AB"(g)$ into $A (g)$ $"A"(g)$ and $B (g)$ $"B"(g)$ in the high temperature limit?