Which of the following is an endothermic reaction?

Calculate $\Delta H$ in $kJ$ for the following reaction:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
Given that:
$H_2O_{(g)} + C_{(s)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H = +131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$

If enthalpies of formation of $C_2H_{4(g)}$,$CO_{2(g)}$ and $H_2O_{(l)}$ at $25 \ ^\circ C$ and $1 \ atm$ pressure are $52$,$-394$ and $-286 \ kJ \ mol^{-1}$ respectively,the enthalpy of combustion of $C_2H_{4(g)}$ will be.....$kJ \ mol^{-1}$.

$C_{(diamond)} + O_2 \to CO_2; \Delta H = -395.3 \ kJ/mole$
$C_{(graphite)} + O_2 \to CO_2; \Delta H = -393.4 \ kJ/mole$
$C_{(graphite)} \to C_{(diamond)}; \Delta H = ?$

The heat of solution of anhydrous $CuSO_4$ and $CuSO_4 \cdot 5 H_2 O$ are $-70 \ kJ \ mol^{-1}$ and $+12 \ kJ \ mol^{-1}$ respectively. The heat of hydration of $CuSO_4$ to $CuSO_4 \cdot 5 H_2 O$ is $-x \ kJ$. The value of $x$ is:

The values of bond enthalpy of $H_2$,$Cl_2$ and $HCl$ are $104$,$58$ and $103 \ kcal/mol$ respectively. Find the enthalpy of formation of $HCl_{(g)}$. Reaction: $\frac{1}{2}H_{2(g)} + \frac{1}{2}Cl_{2(g)} \to HCl_{(g)}$