Reaction of methanol with dioxygen was carried out and Δ H was found to be - 726 kj mol-1 at 298 K. The enthalpy change for the reaction will be CH3OH(l)+(3/2)O2(g) → CO2(g)+2H2O(l); Δ H = - 726 kj mol-1

Question

Reaction of methanol with dioxygen was carried out and Δ H was found to be - 726 kj mol-1 at 298 K. The enthalpy change for the reaction will be CH3OH(l)+(3/2)O2(g) → CO2(g)+2H2O(l); Δ H = - 726 kj mol-1 (A) -741.5 kj mol-1 (B) -727 kj mol-1 (C) +741.5 kj mol-1 (D) +727.2 kj mol-1

Tardigrade Admin · Accepted Answer

Δ ng = 1 - (3/2) =-(1/2) Δ H = Δ U+Δ ngRT = -726× 1000 j mol-1 +(-(1/2)× 8.314 j K-1 mol-1×298 K ) = -727238.78 J mol-1 = - 727 kj mol-1

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

$- 741.5 kj m o l^{- 1}$

$- 727 kj m o l^{- 1}$

$+ 741.5 kj m o l^{- 1}$

$+ 727.2 kj m o l^{- 1}$

$Δ n_{g} = 1 - \frac{3}{2} = - \frac{1}{2}$
$Δ H = Δ U + Δ n_{g} RT$
$= - 726 \times 1000 j m o l^{- 1} + (- \frac{1}{2} \times 8.314 j K^{- 1} m o l^{- 1} \times 298 K)$
$= - 727238.78 J m o l^{- 1} = - 727 kj m o l^{- 1}$

Q. Reaction of methanol with dioxygen was carried out and ΔH was found to be −726kjmol−1 at 298K. The enthalpy change for the reaction will be CH3​OH(l)​+23​O2(g)​→CO2(g)​+2H2​O(l)​; ΔH=−726kjmol−1

−741.5kjmol−1

−727kjmol−1

+741.5kjmol−1

+727.2kjmol−1