Please help?
Calculate deltaHr for the reaction: C3H8(g) + 5O2(g)----->3CO2(g) + 4H2O(l)
Given:
3C(s) + 4H2(g)----->C3H8(g) deltaH -24.8 kcal
H2(g) + 1/2 O2(g)----->H2O(l) deltaH -68.3kcal
C(s) + O2(g)-------> CO2(g) deltaH -94.0 kcal
Let the reactions be numbered 1 to 4 from top to bottom
we need to find deltaHr for reaction 1
reaction 1 can be written as = - reaction 2 +4*reaction
3+ 3* reaction 4
so,
deltaHr (1) = -deltaHr (2) + 4*deltaHr (3) + 3*deltaHr (4)
= - (-24.8) + 4*(-68.3) + 3*(-94)
= 24.8 - 273.2 - 282
= -530.4 Kcal
Answer: -530.4 Kcal
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