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When 0.36 g of glucose was burned in a b...

When 0.36 g of glucose was burned in a bomb calorimeter ( Heat capacity `600 JK^(-1)`)the temperature rise by 10 K. Calculate the standard molar enthalpy of combustion (MJ/mole).

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The correct Answer is:
3
`C_(6)H_(12)O_(6)(S) + 6O^(2)(g) rarr 6CO^(2) (g) + 6H_(2)O(l)`
No. of Mole of `C_(6)H_(12)O_(6) = (0.36)/(180) = 2 xx 10^(-3)`mole
`C=(q)/(DeltaT) " " therefore" " q=C xx Delta T = 600 xx 10 J = 6KJ`
Heat released per mole `=(6)/(2 xx 10^(-3)) = 3 xx 10^(3)KJ =3MJ`
`Delta U =-3ML`
`DeltaH = DeltaU + Deltan_(g)RT" " here " " Delta n_(g)=0`
`Delta H = DeltaU =-3MJ "mole"^(-1)`
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