For complete combustion of methanol CH 3 OH (1)+(3/2) O 2( g ) arrow CO 2( g )+2 H 2 O (1) the amount of heat produced as measured by bomb calorimeter is 726 kJ mol -1 at 27° C. The enthalpy of combustion for the reaction is - x kJ mol -1, where x is . (Nearest integer) (Given: R =8.3 JK -1 mol -1 )

Question

For complete combustion of methanol CH 3 OH (1)+(3/2) O 2( g ) arrow CO 2( g )+2 H 2 O (1) the amount of heat produced as measured by bomb calorimeter is 726 kJ mol -1 at 27° C. The enthalpy of combustion for the reaction is - x kJ mol -1, where x is . (Nearest integer) (Given: R =8.3 JK -1 mol -1 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ U =-726 KJ / mol Δ ng =1-3 / 2=(-1/2) Δ H =Δ U +Δ ngRT =-726-(1/2) × (8.3 × 300/1000) =-727.245

ΔU=−726KJ/mol Δng=1−3/2=2−1​ ΔH=ΔU+ΔngRT =−726−21​×10008.3×300​ =−727.245

$Δ U = - 726 K J / m o l$
$Δ n g = 1 - 3/2 = \frac{- 1}{2}$
$Δ H = Δ U + Δ n g RT$
$= - 726 - \frac{1}{2} \times \frac{8.3 \times 300}{1000}$
$= - 727.245$