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How To Calculate Heat, Work, Energy of A System, Finding Heat Combustion of Propane, and Calculate Heat of Neutralization Reaction of NaOH and HCl

How you can solve these chemistry problems?

1. A system is absorb 35 Joule of heat and has 40 joule work done on it, calculate q, w, and E for this system!
2. Propane, a gas used for cooking in many urban areas, was was placed into a bomb calorimeter with excess oxygen and ignited. When the 0.100 moles of propane is used and the initial temperature of the calorimeter was 25oC and its total heat capacity was 97.1 kj/oC. The reaction raised the temperature of the calorimeter to 27.282oC. Calculate the heat of combustion of propane. The reaction is C3H8(g) + 5O2(g) —? 3CO2(g) + 4H2O(l)
3. When 200 ml of 1.00M NaOH at 25⁰C were mixed with 150 ml of 1.00M HCl, also at 25⁰C is a Styrofoam “coffee-cup calorimeter”, the temperature of the mixture rose to 30⁰C. Calculate the heat of neutralization of NaOH and HCl per mole of water formed. (Assume that the specific heat capacity of the mixture is 4.18 j/gm-oC. Ignore the small heat capacity of the Styrofoam cup)

This is a thermochemical problem and i will show you how you can solve this problem

1. Finding q, w, and E for the system

The system was absorb the heat thus this mean that it is a endothermic process so q = +35 J. Work done on the system thus the system gain energy because of this w = + 40 j, and the total energy of the sistem is:
E = q + w
E = (+35) + (+40)
E = 75 J

What this answer means? This means that the system gain energy from the surrounding, note that the positive sign is indicated that heat is absorb and the work is done on it. When the system release heat or work for the surrounding the sign will be negative.

2. Calculate the heat combustion of propane

first you should calculate the heat that release by the reaction and if this heat is known you can calculate the heat combustion of propane by dividing the heat with propane’s mole.

Increasing temperature
= 27.282 - 25 = 2.282⁰C

Heat evolved by the reaction
= heat that absorb by calorimeter
= Heat capacity x Temperature
= 97.1 KJ/⁰C x 2.282⁰C
= 221.5822 KJ

since the reaction release energy thus the q = ^H = -221.5822 KJ, the sign of the enthalpy in indicated that its a exothermic reaction, when endothermic reaction involved the sign would be positive. To find the heat of combustion of propane we divide this value by the moles of propane

Heat combustion of propane
= -221.5822 KJ / 0.100 mol
= -2215.822 KJ/mol

3. Calculate heat of neutralization of NaOH and HCl reaction

First we must write the reaction and then calculate how many products that produced this calculation involved in applying limitation moles of the reactant. from the reaction we know the ratio between HCl and NaOH is 1:1 thus the moles of HCl is the limit reaction because its the small value when we compare withe moles of NaOH and from the reaction H2O that produce is 150 mmol

moles of NaOH = 200 mL x 1.00 M = 200 mmol
moles of HCl = 150 mL x 1.00 M = 150 mmol
increase of temperature = 30 - 25 = 5oC

the reaction and the stoichiometry;
NaOH + HCl -> NaCl + H2O

	NaOH	+	HCl	->	NaCl	+	H2O
Initial	200		150		150		0
Reacted	150		150		150		150
Equilibrium	50		0		150		150

To calculate the heat that release from the reaction we need the mass of the solution. Total volume of the solution is 350 ml, by assuming that the density of water is 1 g/mL thus the mass of solution would be 350 mL x 1 g/mL = 350 gr. And the heat that release from the reaction is

= specific heat capacity x mass of solution |x increase in temperature
= 4.18 J/g.⁰C x 350 g x 5 ⁰C
= 7315 J

The reaction is exothermic so the enthalpy would be -7315 J, and the thus the heat of neutralization of NaOH and HCl per mole of water formed

= -7315 J, / 0.15 mol
= -48766,7 J/mol
= - 48.8 KJ/mol

Add comment September 8th, 2008