Question

1a)Which of the following processes is exothermic? A) evaporation of morning dew during the day B) the chemical reaction in a ʺhot packʺ often used to treat sore muscles C) conversion of graphite to diamond carbon D) a reaction that abosorbs thermal energy from its surroundings E) reactants converting to higher energy products 1b)What is the change of enthalpy (ΔHrxn) for the following reaction in kJ/mol if 1 mole of carbon reacts with excess oxygen gas to produce carbon monoxide gas and 110.4 kJ of energy is released fro reaction as monitored by an increase in the surrounding temperature. 2 C(s) + O2(g) → CO(g) ΔHrxn = ? A) 220.8 kJ/mol B) -220.8 kJ/mol C) -55.20 kJ/mol D) 110.4 kJ/mol E) -110.4 kJ/mol 1c)Use the ΔH°f information provided to determine ΔH°rxn for the following reaction: ΔH°f (kJ/mol) CH4(g) + 4 Cl2(g) → CCl4(g) + 4 HCl(g) ΔH°rxn = ? CH4(g) -75 CCl4(g) -96 HCl(g) -92 A) -389 kJ B) +79 kJ C) +113 kJ D) -113 kJ E) -71 kJ 1d)Use the standard reaction enthalpies given below to determine ΔH° rxn for the following reaction: 2 SO3(g) + S(s) → 3 SO2(g) ΔH°rxn = ? Given: S(s) + O2(g) → SO2(g) ΔH°rxn = -296.8 kJ 2 SO2(g) + O2(g) → 2 SO3(g) ΔH°rxn = -197.8 kJ A) -494.6 kJ B) 164.9 kJ C) 99.0 kJ D) -164.9 kJ E) 494.6 kJ

Answer 1

Q.1(a):

In evaporation liquid dew is converted to water vapor after receiving heat. Hence this is an endothermic reaction.

Squizing of hot pack induces spontaneous freezing of sodium acetate that creates. Hence (B) is an exothermic reaction.

Graphite is the most stable state of C. Hence graphite has less energy than diamond. Hence heat is absorbed when graphite is converted to diamond. Hence endothermic reaction.

When a reaction absorbs thermal energy it is called endothermic reaction

Since endothermic reaction receives energy, it is converted to a higher energy product..

Q.1(b): Since energy is released, this is an exothermic reaction. This makes the H(rxn) negative.

Hence H(rxn) = - 110.4 kJ/mol (answer)

1.(c): CH4(g) + 4 Cl2(g) → CCl4(g) + 4 HCl(g)

The formula for enthalpy change of a reaction is:

= 1 mol*H_f^o(CCl4,g) + 4 mol*H_f^o(HCl,g) - 1 mol*H_f^o(CH4,g) - 4mol*H_f^o(Cl2,g)

= - 96 kJ + 4*(-92 kJ) - (-75 kJ) - 4*(0 kJ)

= (- 96 - 368 + 75) kJ

= - 389 kJ