Question

Which energy change corresponds to the enthalpy change of atomisation of hydrogen at 298 K?
A the bond energy of a H – H bond
B half the bond energy of a H – H bond
C minus half the bond energy of a H – H bond
D minus the bond energy of a H – H bond

Answer 1

B

Answer 2

could you explain your answer

Answer 3

The reaction characterizing the bond dissociation of Hydrogen gas into its atoms is given as:
\(\rm H_{2(g)} \to 2 H_{(g)}\)

I'm still trying to figure out why "enthalpy of atomization" and "\(standard\) enthalpy of atomization" don't line up for me. But using standard enthalpy of atomization definition:
"enthalpy change when [[1 mol of gaseous atoms is formed]] from its element in its defined physical state under standard conditions (298.15K, 1 atm)."
That part I put in brackets is most important, because to form 1 mol of \(\rm H\) atoms, we are using only half the \( \rm H_2\) due to the ratio in the equation. So that would be half the bond energy which refers to a situation of 1 mol of bonds. And because no new bonds are being formed, the enthalpy is always positive.

I'll keep thinking about this though, since the question doesn't specify the standard pressure, only standard temperature.

Answer 4

I think the answer is B because the enthalpy change of atomisation is +218KJ/mol
and the average bond enthalpy of H-H is 436KJ/mol by halving this number we get 218KJ/mol