The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning water gas obtained from kg of coke is (Assume coke to be 100 % carbon.) (Given enthalpies of combustion of CO2, CO and H2 as 393.5 kJ, 285 kJ, 285 kJ respectively all at 298 K.)

Question

The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning water gas obtained from kg of coke is (Assume coke to be 100 % carbon.) (Given enthalpies of combustion of CO2, CO and H2 as 393.5 kJ, 285 kJ, 285 kJ respectively all at 298 K.) (A) 0.69: 1 (B) 0.96: 1 (C) 0.79: 1 (D) 0.86: 1

Tardigrade Admin · Accepted Answer

Number of moles in 1 kg coke =(1000/12)=83.33 mol (i) For the combustion of 1 kg of coke C + O 2 longrightarrow CO 2 ; Δ H=393.5 kJ ⇒ heat liberated from 1 mole coke =393.5 kJ ∴ heat liberated from 83.33 mole coke =(393.5 × 83.33) kJ (ii) For the burning of water gas C + H 2 O longrightarrow underbrace CO + H 2 text Water gas CO + H 2+ O 2 longrightarrow CO 2+ H 2 O Δ H=285+285=570 kJ ∴ Ratio of heat liberated from burning of water gas obtained from 1 kg of coke H2=(570 × 83.33) kJ ∴ Required ratio = H 1: H 2 =393.5: 570 =0.69: 1

$0.69 : 1$

$0.96 : 1$

$0.79 : 1$

$0.86 : 1$

Q. The ratio of heats liberated at 298K from the combustion of onekg of coke and by burning water gas obtained from kg of coke is (Assume coke to be 100% carbon.) (Given enthalpies of combustion of CO2​, CO and H2​ as 393.5kJ,285kJ,285kJ respectively all at 298K.)

0.69:1

0.96:1

0.79:1

0.86:1

$0.69 : 1$

$0.96 : 1$

$0.79 : 1$

$0.86 : 1$