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Chemistry Question on Enthalpy change

Given ΔHf\Delta H_{f}{ }^{\circ} for CO2(g),CO(g)CO _{2}(g), CO (g) and H2O(g)H _{2} O (g) are 393.5,110.5 and 241.8kJmol1 -393.5, -110.5 \text { and } -241.8 kJ mol ^{-1}, respectively. The ΔHf\Delta H_{f}^{\circ} [in kJmol1kJ mol ^{-1} ] for the reaction CO2(g)+H2(g)CO(g)+H2O(g)CO _{2}(g)+ H _{2}(g) \longrightarrow CO (g)+ H _{2} O (g) is

A

524.1

B

-262.5

C

-41.7

D

41.2

Answer

41.2

Explanation

Solution

CO2(g)+H2(g)CO(g)+H2O(g)CO _{2}(g)+ H _{2}(g) \longrightarrow CO (g)+ H _{2} O (g)
ΔHf=(ΔH)products (ΔH)reactants \Delta H_{f}^{\circ}=(\Delta H)_{\text {products }}-(\Delta H)_{\text {reactants }}
Hence, ΔHf=[ΔHfCO+ΔHfH2O][ΔHfCO2](i)\Delta H_{f}{ }^{\circ}=\left[\Delta H_{f_{ CO }}+\Delta H_{f_{ H _{2} O }}\right]-\left[\Delta H_{f CO _{2}}\right] \ldots( i )
Given, ΔHfCO2(s)=393.5kJmol1\Delta H_{f CO _{2}(s)}^{\circ}=-393.5 kJ mol ^{-1}
ΔHfCO=110.5kJmol1\Delta H_{f CO }^{\circ}=-110.5 \,kJ mol ^{-1}
ΔHfH2O=241.8kJmol1\Delta H_{f H _{2} O }^{\circ}=-241.8 kJ mol ^{-1}
Put there value in equation
ΔHf=110.5241.8+393.5=41.2\Delta H_{f^{\circ}}=-110.5-241.8+393.5=41.2