i)
The balanced equation is "\\ C_6H_5COOH(s) +\\dfrac{15}{2} O_2(g) \u2192 7CO_2(g) +3 H_2O(l)"
ii)
"Molecular \\ mass\\ of\\ benzoic\\ acid=12*6+5+12+16*2+1=122 g"
as 1 g produces 24.6 kJ of heat so heat produced by one mole of benzoic acid is equal to
=24.6*122 =3001.2 kJ which is the "\\Delta H"
iii) Heat released from 2.25 g of benzoic acid = 2.25*24.6 = 55.35 kJ
iv) change in heat = ms (T2-T1)
55.35*103 =500*4.18*(T2-20)
T2 =46.5oC
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