An intimate mixture of ferric oxide and aluminium is used as solid fuel in rockets. Calculate the fuel value per `cm^(3)` of the mixture. Heats of formation and densities are as follows:
`H_(f(Al_(2)O_(3)))=-399kcal” “mol^(-1),H_(f(Fe_(2)O_(3)))=-199kcal” “mol^(-1)` ltbr. Density of `Fe_(2)O_(3)=5.2g//cm^(3),” Density of “Al=2.7g//cm^(3)`
`H_(f(Al_(2)O_(3)))=-399kcal” “mol^(-1),H_(f(Fe_(2)O_(3)))=-199kcal” “mol^(-1)` ltbr. Density of `Fe_(2)O_(3)=5.2g//cm^(3),” Density of “Al=2.7g//cm^(3)`
Correct Answer – `8 jk^(-1)MIL^(-1)`
The required equationis: `2Al + Fe_(2)O_(3) rarr Al_(2)O_(3) + 2Fe, Delta H = ?`
`Delta H = Delta H_(f(“products”)) – Delta H_(f (“reac tan is”))`
`[Delta H_(f(Al_(2)O_(3))) + 2 Delta H_(f(Fe))] – [2 DeltaH_(f (Al)) + DeltaH_(f (Fe_(2)O_(3)))]`
`= (0399 + 2 xx 0) – [2 xx 0 + (-195.92)]`
`=-399 + 195.92 = – 203.08`
At, mass of aluminium `=27`, Mol. mass of `Fe_(2)O_(3) = 160`
Volume of reacts `(160)/(5.2) + (2 xx 27)/(2.7) = 50.77 cm^(3)`
Fuel value per `cm^(3) = (-203.08)/(50.77) = 4`k cal.