   Chapter 5, Problem 120SCQ

Chapter
Section
Textbook Problem

In the reaction of two moles of gaseous hydrogen and one mole of gaseous oxygen to form two moles of gaseous water vapor, two moles of products are formed from three moles of reactants. If this reaction is done at 1.01 × 104 Pa (and at 0 °C), the volume is reduced by 22.4 L. (a) In this reaction, how much work is done on the system (H2, O2, H2O) by the surroundings? (b) The enthalpy change for this reaction is −483.6 kJ. Use this value, along with the answer to (a), to calculate ΔrU, the change in internal energy in the system.

(a)

Interpretation Introduction

Interpretation:

The work done on the system by the surroundings has to be determined

Concept Introduction:

The standard enthalpy of formation is the enthalpy change for the formation of 1mol of the compound directly from its component elements in their standard states.

Enthalpy change for the reaction ΔrH0 ΣnΔfH0(products)-ΣnΔfH0(reactants)

Explanation

Given,

Volume=22.4L

Pressure=1.01×105Pa

Substitute in W =-P(ΔV) as,

W=-(22.4L×1

(b)

Interpretation Introduction

Interpretation:

The change in internal energy in the system has to be calculated.

Concept Introduction:

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 change for the reaction ΔrH0=ΣnΔfH0(products)-ΣnΔfH0(reactants)

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