6C( graphite )+3H2( g)→C6H6(ℓ);ΔH=48.5 kJ/molC( graphite )+O2( g)→CO2( g);ΔH=−393.5 kJ/molH2(g)+21( g)⟶H2O(ℓ);ΔH=−286 kJ/mol equation −(1)×1+(2)×6+(3)×3−48.5−6×393.5−3×286=−3267.5 kJ for 1 mol=−6535 kJ for 2 mol Ans. 6535 kJ
Combustion of 1 mole of benzene is expressed at C_6 H_6( )+ 15 2 O_2( ~g) → 6 CO_2( ~g)+3 H_2 O( ) . The standard enthalpy of combustion of 2 ~mol of…
JEE Main 2024 — Chemistry Physical Chemistry
2024integermedium
Combustion of 1 mole of benzene is expressed at C6H6(l)+215O2( g)→6CO2( g)+3H2O(l). The standard enthalpy of combustion of 2 mol of benzene is −′x′kJ. x= ______ Given: 1. standard Enthalpy of formation of 1 mol of C6H6(l), for the reaction 6C (graphite) +3H2( g)→C6H6(l) is 48.5 kJ mol−1. 2. Standard Enthalpy of formation of 1 mol of CO2( g), for the reaction C (graphite) +O2( g)→CO2( g) is −393.5 kJ mol−1. 3. Standard and Enthalpy of formation of 1 mol of H2O(l), for the reaction H2( g)+21O2( g)→H2O(l) is −286 kJ mol−1.
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