In the reaction for the transition of carbon in the diamond form to carbon in the graphite form,$\Delta H = -453.5 \ cal$. This points out that

Which of the following is an endothermic reaction?

The heat of neutralization of a weak acid and a strong base is less than the heat of neutralization of a strong acid and a strong base due to:

What is the enthalpy change when $6.80 \, g$ of $NH_3$ is passed over hot $CuO$ (in $, kJ$)? The standard enthalpies of formation for $NH_3(g)$,$CuO(s)$,and $H_2O(l)$ are $-46.0$,$-155.0$,and $-285.0 \, kJ \, mol^{-1}$ respectively. The reaction is: $NH_3(g) + \frac{3}{2}CuO(s) \to \frac{1}{2}N_2(g) + \frac{3}{2}H_2O(l) + \frac{3}{2}Cu(s)$

At $298 \, K$,the bond energies of $C-H$,$C-C$,$C=C$,and $H-H$ bonds are $414$,$347$,$615$,and $435 \, kJ \, mol^{-1}$ respectively. What is the enthalpy change for the reaction $H_2C=CH_{2(g)} + H_{2(g)} \rightarrow H_3C-CH_{3(g)}$ at $298 \, K$?

Two reactions are given below:
$2 Fe_{(s)} + \frac{3}{2} O_{2_{(g)}} \rightarrow Fe_2 O_{3_{(s)}}, \Delta H^{o} = -822 \ kJ/mol$
$C_{(s)} + \frac{1}{2} O_{2_{(g)}} \rightarrow CO_{(g)}, \Delta H^{o} = -110 \ kJ/mol$
Then,the enthalpy change for the following reaction is:
$3 C_{(s)} + Fe_2 O_{3_{(s)}} \rightarrow 2 Fe_{(s)} + 3 CO_{(g)}$