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Standard XII

Chemistry

Heat of Reaction

Reaction of m...

Question

Reaction of methanol with dioxygen was carried out and $Δ U$ was found to be - 726 kJ $m o l^{- 1}$ at 298 K. The enthalpy change for the reaction will be:
$C H_{3} O H (l) + \frac{3}{2} O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(l)} Δ H = - 726 K J m o l^{(- 1)}$

- 741.5 K J m o l^{(- 1)}

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- 727 K J m o l^{(- 1)}

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+ 741.5 K J m o l^{(- 1)}

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+ 727 K J m o l^{(- 1)}

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Solution

The correct option is B $- 727 K J m o l^{(- 1)}$
Enthalpy change for reaction $(Δ H)$ is given by the expression
$Δ H = Δ U + Δ n_{g} R T$
where,
$Δ H =$ Change in internal energy
$Δ n_{g} =$ Change in number of moles
For the given reaction,
$Δ n_{g} = \sum n_{g} (p r o d u c t) - \sum n_{g} (r e a c t a n t s)$
$= 1 - \frac{3}{2}$
$Δ n_{g} = - 0.5 m o l e s$
$Δ U = - 726 k J$
$T + 298 K$
$R = 8.314 \times 10^{- 3} k J / K m o l$
Substituting the values
$Δ H = (- 726) + (- 0.5 \times 298 \times 8.314 \times 10^{- 3})$
$Δ H = - 726 - 1.23$
$Δ H = - 727 k J / m o l$

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Reaction of methanol with dioxygen was carried out and ΔU was found to be - 726 kJ mol−1 at 298 K. The enthalpy change for the reaction will be:CH3OH(l)+32O2(g)→CO2(g)+2H2O(l) ΔH=−726KJmol(−1)

Reaction of methanol with dioxygen was carried out and $Δ U$ was found to be - 726 kJ $m o l^{- 1}$ at 298 K. The enthalpy change for the reaction will be:
$C H_{3} O H (l) + \frac{3}{2} O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(l)} Δ H = - 726 K J m o l^{(- 1)}$