Calculate the standard enthalpy change of the following reaction:
$C_2H_{4(g)} + 3O_{2(g)} \longrightarrow 2CO_{2(g)} + 2H_2O_{(\ell)}$
Given:
$\Delta_{f}H^{\circ}(C_2H_4) = 52 \ kJ \ mol^{-1}$
$\Delta_{f}H^{\circ}(CO_2) = -393.5 \ kJ \ mol^{-1}$
$\Delta_{f}H^{\circ}(H_2O) = -285.8 \ kJ \ mol^{-1}$
(Note: Standard values adjusted for accuracy)
$C_2H_{4(g)} + 3O_{2(g)} \longrightarrow 2CO_{2(g)} + 2H_2O_{(\ell)}$
Given:
$\Delta_{f}H^{\circ}(C_2H_4) = 52 \ kJ \ mol^{-1}$
$\Delta_{f}H^{\circ}(CO_2) = -393.5 \ kJ \ mol^{-1}$
$\Delta_{f}H^{\circ}(H_2O) = -285.8 \ kJ \ mol^{-1}$
(Note: Standard values adjusted for accuracy)
- A$-1411.1 \ kJ \ mol^{-1}$
- B$-1300 \ kJ \ mol^{-1}$
- C$-1950 \ kJ \ mol^{-1}$
- D$-1500 \ kJ \ mol^{-1}$
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View SolutionCalculate the heat of formation of $Ca(OH)_{2(s)}$ at $1.8\,^{\circ}C$ from the following data:
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View SolutionOn the basis of the thermochemical equations:
$H_{2}O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} \quad \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} \quad \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_{2}O_{(g)} \quad \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = X \ kJ$
The value of $X$ will be $.... \ kJ$.
$H_{2}O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} \quad \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} \quad \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_{2}O_{(g)} \quad \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = X \ kJ$
The value of $X$ will be $.... \ kJ$.
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