Calculate the standard enthalpy change of the reaction: $C_2H_{2(g)} + \frac{5}{2}O_{2(g)} \rightarrow 2CO_{2(g)} + H_2O_{(\ell)}$ given the following standard enthalpies of formation:
$\Delta_fH^{\circ}(CO_2) = -393 \ kJ \ mol^{-1}$
$\Delta_fH^{\circ}(H_2O) = -286 \ kJ \ mol^{-1}$
$\Delta_fH^{\circ}(C_2H_2) = 227 \ kJ \ mol^{-1}$

  • A
    $-650 \ kJ \ mol^{-1}$
  • B
    $-1950 \ kJ \ mol^{-1}$
  • C
    $-1299 \ kJ \ mol^{-1}$
  • D
    $-2598 \ kJ \ mol^{-1}$

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Similar Questions

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

Given the reactions:
$C + \frac{1}{2}O_2 \to CO : \Delta H = -12 \ kJ$
$CO + \frac{1}{2}O_2 \to CO_2 : \Delta H = -10 \ kJ$
For the reaction $C + O_2 \to CO_2 : \Delta H = x \ kJ$,the value of $x$ is: (in $kJ$)

The heat of neutralization of a strong acid by a strong base is nearly equal to

The heat of combustion of carbon is $-94 \ kcal$ at $1 \ atm$ pressure. The internal energy of $CO_2$ is .... $kcal$.

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

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