Question 27 of 35SubmitSubstanceAH (kJ/mol)O2 (g)SO2 (g)-296.9PbO (s)-217.3kJ/mol14C7+/-x 10 0Tap here or pull up for additional resourcesLO00 Question 27 of 35SubmitUsing the provided table and theequation below, determine the heatof formation for PbS.2 PbS (s) + 3 O2 (g) → 2 SO2 (g) + 2PbO (s) AH° = -828.4 kJ/molSubstanceAH? (kJ/mol)O2 (g)kJ/mol134C7+/-x 10 0Tap here or pull up for additional resourcesLO00

Answered: Using the provided table and the…

Chemistry by OpenStax (2015-05-04)

1st Edition

ISBN:9781938168390

Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Chapter5: Thermochemistry

Section: Chapter Questions

Problem 17E: Would the amount of heat absorbed by the dissolution in Example 5.6 appear greater, lesser, or...

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Question 27 of 35
Submit
Using the provided table and the
equation below, determine the heat
of formation for PbS.
2 PbS (s) + 3 O2 (g) → 2 SO2 (g) + 2
PbO (s) AH° = -828.4 kJ/mol
Substance
AH? (kJ/mol)
O2 (g)
kJ/mol
1
3
4
C
7
+/-
x 10 0
Tap here or pull up for additional resources
LO
00

Expert Solution

Introduction

The enthalpy of formation is the change in enthalpy on formation of one mole of substance from its elements under standard conditions of 1 atm pressure and 298.15 K temperature.

It accounts for the energy consumed or released in order to form one mole of substance from its elements.

Solution

Equation of the reaction

$2 {PbS}_{(s)} + 3 O_{2_{(g)}} \to 2 {SO}_{2_{(g)}} + 2 {PbO}_{(s)}$

Data provided:

$∆ H^{°} = - 828.4 kJ / mol ∆ {H_{f}}^{°} O_{2} = 0 kJ / mol ∆ {H_{f}}^{°} {SO}_{2} = - 296.9 kJ / mol ∆ {H_{f}}^{°} PbO = - 217.3 kJ / mol$

Standard ethaply change of formation is given as:

$∆ {H^{°}}_{reaction} = (sum of ∆ {H^{°}}_{products}) - (sum of ∆ {H^{°}}_{reactants})$

Putting the values in the equation

$- 828.4 = (2 \times - 296.9 + 2 \times - 217.3) - (2 \times 2 PbS + 0) - 828.4 = (- 593.8 - 434.6) - 2 PbS - 828.4 + 1028.4 = - 2 PbS Pbs = - 100 kJ / mol$

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