The heat of solution of $Na_2SO_{4(s)}$ and $Na_2SO_4 \cdot 10H_2O_{(s)}$ are $-2.34 \ kJ \ mol^{-1}$ and $78.87 \ kJ \ mol^{-1}$ respectively. The heat of hydration of $Na_2SO_{4(s)}$ is:

The heat of formation of methane $C_{(s)} + 2H_{2(g)} \to CH_{4(g)}$ at constant pressure is $-18500 \ cal$ at $25 \ ^oC$. The heat of reaction at constant volume would be (in $cal$)

Calculate the enthalpy of formation of sucrose $(C_{12}H_{22}O_{11})$ in $kJ \, mol^{-1}$ using the following data:
$(i) \, C_{12}H_{22}O_{11} + 12O_2 \to 12CO_2 + 11H_2O, \, \Delta H = -5200.7 \, kJ \, mol^{-1}$
$(ii) \, C + O_2 \to CO_2, \, \Delta H = -394.5 \, kJ \, mol^{-1}$
$(iii) \, H_2 + \frac{1}{2}O_2 \to H_2O, \, \Delta H = -285.8 \, kJ \, mol^{-1}$

Given,
$NO_{(g)} + O_{3(g)} \longrightarrow NO_{2(g)} + O_{2(g)}; \Delta H = -198.9 \, kJ/mol$
$O_{3(g)} \longrightarrow 3/2 O_{2(g)}; \Delta H = -142.3 \, kJ/mol$
$O_{2(g)} \longrightarrow 2O_{(g)}; \Delta H = +495.0 \, kJ/mol$
The enthalpy change $(\Delta H)$ for the following reaction is $..... \, kJ/mol$
$NO_{(g)} + O_{(g)} \longrightarrow NO_{2(g)}$

Which of the following statements is correct?

Find the value of $Q$ from the following equations:
$(i)$ $C_{(s)} + O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = Q \ kJ$
$(ii)$ $C_{(s)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{(g)}$ $\Delta H = -x \ kJ$
$(iii)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = -y \ kJ$