The values of Δ r H ° and Δ r U ° at 298 K has to be calculated for the given reaction. Concept introduction: Enthalpy is the internal heat content of the system. Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system. Therefore, the measure of amount of heat supplied to the system gives the enthalpy change of the system. The total amount of energy possessed within the given system is known as internal energy of the system. It is denoted by U .

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Definition Definition Number that is expressed before molecules, ions, and atoms such that it balances out the number of components present on either section of the equation in a chemical reaction. Stoichiometric coefficients can be a fraction or a whole number and are useful in determining the mole ratio among the reactants and products. In any equalized chemical equation, the number of components on either side of the equation will be the same.

Chapter 2, Problem 2C.7AE

(i)

Interpretation Introduction

Interpretation: The values of $ΔrH°$ and $ΔrU°$ at $298 K$ has to be calculated for the given reaction.

Concept introduction: Enthalpy is the internal heat content of the system. Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system. Therefore, the measure of amount of heat supplied to the system gives the enthalpy change of the system. The total amount of energy possessed within the given system is known as internal energy of the system. It is denoted by $U$ .

(i)

Expert Solution

Answer to Problem 2C.7AE

The value of $ΔrH°$ for the reaction is $131.29 kJ mol_−1$ . The value of $ΔrU°$ for the reaction is $128.81kJ mol-1_$ .

Explanation of Solution

The given reaction is:

$C(graphite) + H2O(g) → CO(g) + H2(g)$ (1)

The standard enthalpy for the reaction is calculated by the formula,

$ΔrHο=∑m⋅ΔHfο(products)−∑n⋅ΔHfο(reactants)$ (2)

Where,

• $ΔHfο(products)$ is the standard enthalpy of formation of products.
• $ΔHfο(reactants)$ is the standard enthalpy of formation of reactants.
• $m$ is the stoichiometric coefficient of products.
• $n$ is the stoichiometric coefficient of reactants

For standard enthalpy of the reaction, equation (2) is written as,

$ΔrHο=(1×ΔHfο(CO)+1×ΔHfο(H2)−(1×ΔHfο(C(graphite))+1×ΔHfο(H2O)))$

The standard enthalpy of formation of $CO$ is $−110.53 kJ mol−1$ .

The standard enthalpy of formation of $H2$ is $0 kJ mol−1$ .

The standard enthalpy of formation of $H2O$ is $−241.82 kJ mol−1$ .

The standard enthalpy of formation of $C(graphite)$ is $0 kJ mol−1$ .

Substitute the values of standard enthalpy of formation of reactants and products in the above equation to calculate the standard enthalpy of the reaction.

$ΔrHο=(1×(−110.53 kJ mol−1)+1×(0 kJ mol−1)−(1×(0 kJ mol−1)+1×(−241.82 kJ mol−1)))=−110.53 kJ mol−1+241.82 kJ mol−1=131.29 kJ mol_−1$

Hence, the value of $ΔrH°$ for the reaction is $131.29 kJ mol_−1$ .

The change in number of moles of the reaction is calculated by the formula,

$Number of moles, Δng= Moles of gaseous product− Moles of gaseous reactant$

Hence, the number of moles for the reaction is,

$Δng= (1+1)−(1)=1$

The expression to calculate the value of $ΔrU°$ for the reaction is,

$ΔrU°= ΔrH°−ΔngRT$

Substitute the values of $ΔrH°$ , $R$ , $Δng$ and $T= 298 K$ in the above equation to calculate the value of $ΔrU°$ .

$ΔrU°= 131.29 kJ mol−1 −(1×8.314× 10−3 kJ K−1 mol−1×298 K)= 131.29 kJ mol−1 − (2477.572× 10−3kJ mol−1)= 128.81kJ mol-1_$

Hence, the value of $ΔrU°$ for the reaction is $128.81kJ mol-1_$ .

(ii)