Question

The compound heptane, C7H16, is a good fuel. It is a liquid at ordinary temperatures. When the liquid is burned, the reaction involved is C7H16(ℓ) + 11 O2(g)7 CO2(g) + 8 H2O(g)

The standard enthalpy of formation of liquid heptane at 25 °C is -224.2 kJ mol-1; other relevant enthalpy of formation values in kJ mol-1 are: C7H16(g) = -187.6 ; CO2(g) = -393.5 ; H2O(g) = -241.8

(a) Calculate the enthalpy change in the burning of 2.000 mol liquid heptane to form gaseous products at 25°C.

State explicitly whether the reaction is endothermic or exothermic. ΔH° = kJ (b) Would more or less heat be evolved if gaseous heptane were burned under the same conditions?

What is the standard enthalpy change for vaporizing 2.000 mol C7H16() at 25°C? ΔH° = kJ

Calculate the enthalpy change in the burning of 2.000 mol gaseous heptane to form gaseous products at 25°C. ΔH° = kJ

Answer 1

The siaantion Answer D means Imale he ptane soact with I mole o O togive t hmple o co and mabt of HeO So EAH neartant SΔΗροue! 7 xCCo)RCH, D)-ixltH)+lLx l0,) x-19348x(-2419)-[1x 24 2) +11 0 -2154 5t(-1934 4) + 224 2 -4464-7 kI Mol 2 moles af theptane 2X-4464 7 AHstadeFron 1 89 29-4 1 Hente exotherimic because AH The siyn will be us negative l 2 moles af gase oue heptane CueTe buined under the Same CondAoo Finswes Cb) 4Hsaatron 1x (-393.54&241-3)- 1x = 2154 - 1934 t87.6 - 4501-3 KT Mol- )- () 2 mols 2X-4501.3 fan -9002 6 KJ Heat invalved s mase than that hoat invalved in liquid