At 25° C and 1 atm pressure, the enthalpy of combustion of benzene (1) and acetylene ( g ) are -3268 kJ mol -1 and -1300 kJ mol -1, respectively. The change in enthalpy for the reaction 3 C 2 H 2( g ) arrow C 6 H 6( l ), is

Question

At 25° C and 1 atm pressure, the enthalpy of combustion of benzene (1) and acetylene ( g ) are -3268 kJ mol -1 and -1300 kJ mol -1, respectively. The change in enthalpy for the reaction 3 C 2 H 2( g ) arrow C 6 H 6( l ), is (A) +324 kJ mol -1 (B) +632 kJ mol -1 (C) -632 kJ mol -1 (D) -732 kJ mol -1

Tardigrade Admin · Accepted Answer

Δ H =∑ Δ H text Combustion ( Reactant )-∑ Δ H text Combustion (Product) =3 ×(-1300)-[-3268] =-632 kJ mol -1

Q. At $2 5^{\circ} C$ and $1$ atm pressure, the enthalpy of combustion of benzene (1) and acetylene ( $g)$ are $- 3268 k J m o l^{- 1}$ and $- 1300 k J m o l^{- 1}$ , respectively. The change in enthalpy for the reaction $3 C_{2} H_{2} (g) \to C_{6} H_{6} (l)$ , is

$+ 324 k J m o l^{- 1}$

$+ 632 k J m o l^{- 1}$

$- 632 k J m o l^{- 1}$

$- 732 k J m o l^{- 1}$

$Δ H = \sum Δ H_{Combustion} ($ Reactant $) - \sum Δ H_{Combustion}$ (Product)
$= 3 \times (- 1300) - [- 3268]$
$= - 632 k J m o l^{- 1}$

Q. At 25∘C and 1 atm pressure, the enthalpy of combustion of benzene (1) and acetylene ( g) are −3268kJmol−1 and −1300kJmol−1, respectively. The change in enthalpy for the reaction 3C2​H2​(g)→C6​H6​(l), is

+324kJmol−1

+632kJmol−1

−632kJmol−1

−732kJmol−1