   Chapter 5.6, Problem 5.7CYU

Chapter
Section
Textbook Problem

Assume 200. mL of 0.400 M HCl is mixed with 200. mL of 0.400 M NaOH in a coffee-cup calorimeter The temperature of the solutions before mixing was 25.10 °C; after mixing and allowing the reaction to occur, the temperature is 27.78 °C. What is the enthalpy change when one mole of acid is neutralized? (Assume that the densities of all solutions are 1.00 g/mL and their specific heat capacities are 4.20 J/g · K.)

Interpretation Introduction

Interpretation:

The enthalpy change for the given reaction has to be calculated.

Concept Introduction:

Standard enthalpy

Standard enthalpy of the reaction,ΔrHo, is the change in enthalpy that happens when matter is transformed by a given chemical reaction, when all reactants and products are in their standard states.

Enthalpy of the reaction,ΔrH, is the change in enthalpy that happens when matter is transformed by a given chemical reaction

Heat energy required to raise the temperature of 1g of substance by 1K. Energy gained or lost can be calculated using the below equation.

q=C×m×ΔT

Where,

q= energy gained or lost for a given mass of substance (m)

C =specific heat capacity,

ΔT= change in temperature.

The change in enthalpy, ΔH in kJ per mole of a given reactant for the reaction can be calculated as:

ΔrH=enthalpy changenumber of moles

ΔrH=ΔHnumber of moles

Explanation

Given,

Specific heat capacity of the solution =4.20JK/g

Density of HCl=1g/mL

Mass of HCl=1g/mL× 200mL =200g

Mass of NaOH = 1g/mL×200mL=200g

Mass of solution =400g

ΔT= 27.780C-25.100C=2.680C=2.68K

Amount of HCl=Amount of NaOH=Concentration × Volume=0.400mol/L×0.200mL

Amount of HCl=Amount of NaOH=0.0800mol

Assume qr+qsol=0

qsol Can be calculated from q=C×m×ΔT, as

qsol=400g×4

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