I think nothing happens (At standard conditions) as there is insufficient heat in the atmosphere to allow breakage of bonds in the reactants. Hence one must use Hess's Law and enthalpy changes of combustion for the products and reactants (already calculated and given to student) to calculate the enthalpy change of reaction for C and H2.
I think nothing happens (At standard conditions) as there is insufficient heat in the atmosphere to allow breakage of bonds in the reactants. Hence one must use Hess's Law and enthalpy changes of combustion for the products and reactants (already calculated and given to student) to calculate the enthalpy change of reaction for C and H2.
Is that correct?
They simply don't react (as you say) This is because the activation energy is so high (as you allude to) Which makes any rate of reaction very slow (which means any energy change would not be detectable since any heat gain/loss by the reaction would be lost to/gained from the surroundings before it could be detected) And anyway, even if C did react with H2, what is the chance that 1 mol of C and 2 mol of H2 would form 1 mol of CH4 and nothing else, e.g. ethane, benzene etc. etc.