Question

The compound 1-pentanol, C5H12O, is a good fuel. It is a liquid at ordinary temperatures. When the liquid is burned, the reaction involved is

2 C5H12O(ℓ) + 15 O2(g)10 CO2(g) + 12 H2O(g)

The standard enthalpy of formation of liquid 1-pentanol at 25 °C is -351.6 kJ mol-1; other relevant enthalpy of formation values in kJ mol-1 are:
C5H12O(g) = -294.6 ; CO2(g) = -393.5 ; H2O(g) = -241.8

(a) Calculate the enthalpy change in the burning of 5.000 mol liquid 1-pentanol 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 1-pentanol were burned under the same conditions?

What is the standard enthalpy change for vaporizing 5.000 mol C5H12O() at 25°C?

ΔH° =
kJ

Calculate the enthalpy change in the burning of 5.000 mol gaseous 1-pentanol to form gaseous products at 25°C.

ΔH° =
kJ

Answer 1

E5€ +6_+O_= 64,0(1) -351-6 kJ/mol o n C+02 - COLG) = -393-5 kJ/mol - © H+ 0. = 4,06) D#- 241-8 kg/mol - © Multiplying equation ọ by (2) equation ③ by 10 most an equation ③ by 12 and summing up all the equations, we get- 2654,08% + 1509)= 1000_CW) + 124,063) of According to bless's Low, Alting for this Reaction is - = (223516) -(104393-5) =(22311-3) = [-613304 kJ/mol So, when 5 ml liquicl -Pantanal is busent, sha= 5* 461334) kot 2-30667 ko :: BHP-30667 kJ Since the Sign of sth is negative , at follows that the reaction is EXOTHERMICO b) Se+64+<Q=C4,0 6) 0–2946kt/mol - Similarly by multitlying equation s by 62), equation © by 19, equation © by 12 and sunning up, we get 2G H2O(g) + 150, 18) = 10 CO, (g) + 2Holl 306

This Mal According to Hear's Law, Alt for this Reaction is a =(b) – frox393-5)-(124240-8) = 6247-4 kJ/mol under the same Marex Hent i, enabled when gewees 1-Partond is burent conditions. The AH® for Vaporizing s hol 634,061) at ase = 54 55 5476) - = 5*(8516-24-6). (s«=285 kt » Dt for burning 5 ml of gaseous Cg H₂O is a +5+ (626~) = 431237 k 4.1°==31237 ko fire burning 25 CE 5 SX 351 .4 gaslacus