Calculate ΔH°R ,1450 K for the reaction 1/2H2(g) + 1/2Cl2(g) →HCl(g) and 1 bar pressure given that ΔH°f(HCl,g)=-92.3 KJ mol-¹ at 298.15 K and that Cp,m(H2,g)=(29.064-0.8363 × 10-³ TK +20.111× 10^-7 T²K²)JK-¹ mol-¹ Cp,m(Cl2,g)=(31.695+10.143 × 10-³ TK -40.373× 10^-7 T²K²)JK-¹ mol-¹

Question

Calculate [latex]\Delta\mathrm{H°}_{R,1450 K}[/latex] for the reaction [latex]1/2 H_{2}\left( g \right)+1/2 Cl_{2}\left( g \right)\longrightarrow HCl\left( g \right)[/latex] and 1 bar pressure given that [latex]\Delta H°_{f}\left( HCl,g \right)=-92.3 KJ mol^{-1}[/latex] at 298.15 K and that

[latex]C_{P,m}\left( H_{2},g \right)=\left( 29.064 -0.836 ×10^{-3}\frac{T}{K}+20.111×10^{-7}\frac{T^{2}}{K^{2}} \right) J K^{-1}mol^{-1}[/latex]

[latex]C_{P,m}\left( Cl_{2},g \right)=\left( 31.695 +10.143×10^{-3}\frac{T}{K}-40.373×10^{-7}\frac{T^{2}}{K^{2}} \right) J K^{-1}mol^{-1}[/latex]

[latex]C_{P,m}\left( HCl_{2},g \right)=\left( 28.165 +1.809 ×10^{-3}\frac{T}{K}+15.464×10^{-7}\frac{T^{2}}{K^{2}} \right) J K^{-1}mol^{-1}[/latex]

over this temperature range. The ratios T/K and [latex]T^{2}/K^{2}[/latex] appear in these equations in order to have the right units for the heat capacity.

Accepted Answer

[latex]\Delta H°_{R,1450 K}=\Delta H°_{R,298.15 k}+\int_{298.15}^{1450}\Delta C_{P}\left( T \right)dT[/latex]

[latex]\Delta C_{P}\left( T \right)=\left[ 28.165 +1.809 × 10^{-3} \frac{T}{k}+15.464 × 10^{-7} \frac{T^{2}}{K^{2}}\right][/latex]

[latex]-\frac{1}{2}\left( 29.064 -0.8363 × 10^{-3} \frac{T}{K}+20.111×10^{-7}\frac{T^{2}}{K^{2}}\right)[/latex]

[latex]-\frac{1}{2}\left( 31.695 +10.143 × 10^{-3} \frac{T}{K}-40.373 ×10^{-7}\frac{T^{2}}{K^{2}}\right)] JK^{-1} mol^{-1} [/latex]

[latex] =\left( -2.215 -2.844 × 10^{-3}\frac{T}{K}+25.595 ×10^{-7}\frac{T^{2}}{K^{2}} \right) JK^{-1} mol^{-1}[/latex]

[latex]\Delta H°_{R,1450 K}=-92.3 KJ mol^{-1}[/latex]

[latex]+\int_{298.15}^{1450}\left( -2.215-2.844 × 10^{-3}\frac{T}{K}+25.595 ×10^{-7}\frac{T^{2}}{K^{2}} \right) × d\frac{T}{K}J mol^{-1} [/latex]

[latex]=-92.3 KJ mol^{-1} -2.836 KJ mol^{-1}=-95.1 KJ mol^{-1}[/latex]

In this particular case, the change in the reaction enthalpy with T is not large. This is
the case because [latex]\Delta C_{P}\left( T \right)[/latex] is small and not because an individual [latex]C_{P,i}\left( T \right)[/latex] is small.

Question 4.2: Calculate ΔH°R ,1450 K for the reaction 1/2H2(g) + 1/2Cl2(g)...

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