Question

the standard enthalpy change for the following rea is 436.4 kJ/mol :H2(g)--> H(g) +H(g) . calc the standard enthalpy of formatiom of atomic hydrogen

Answer 1

I think your answer is

_____Thermo

H2g) -----> 2H(g)____dHrxn = 436.4 kJ/mole H2 OR 436.4 kJ/2 mole H

dHrxn = 2*dHf H - dHf H2 = 2*dHf H as dHf H2 =0

so, dHf H = ??

Answers

436.4/2=218.2