Chemistry, 30.07.2019 02:00 shauna94

For which of the following reactions is δh∘rxn equal to δh∘f of the product(s)? you do not need to look up any values to answer this question. check all that apply. check all that apply. c(s, graphite)+o2(g)→co2(g) li(s)+12f2(l)→lif(s) co(g)+12o2(g)→co2(g) baco3(s)→bao(s)+co2(g) 2li(s)+f2(g)→2lif(s) li(s)+12f2(g)→lif(s)

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Answer from: conniehodson

Answer : The reactions are,

$C(graphite)+O_2(g)\rightarrow CO_2(g)$

$BaCO_3(s)\rightarrow BaO(s)+CO_2(g)$

Explanation :

The general expression for standard enthalpy of reaction will be,

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

The standard enthalpy of formation of solid and liquid always zero and standard enthalpy of every element in its standard state is equal to zero.

Now we have to determine the standard enthalpy of reaction of the following reactions.

(1) $C(graphite)+O_2(g)\rightarrow CO_2(g)$

$\Delta H^o_{rxn}=[n_{CO_2}\times \Delta H_f(CO_2)]-[(n_{O_2}\times \Delta H_f(O_2))+(n_f(C)\times \Delta H_f(C))]\\\\\Delta H^o_{rxn}=\Delta H_f(CO_2)-[0+0]=H_f(CO_2)$

(2) $BaCO_3(s)\rightarrow BaO(s)+CO_2(g)$

$\Delta H^o_{rxn}=H_f(CO_2)$

(3) $CO(g)+12O_2(g)\rightarrow CO_2(g)$

$\Delta H^o_{rxn}=[n_{CO_2}\times \Delta H_f(CO_2)]-[(n_{CO}\times \Delta H_f(CO))]\\\\\Delta H^o_{rxn}=[\Delta H_f(CO_2)]-[\Delta H_f(CO)]$

(4) $Li(s)+12F_2(g)\rightarrow LiF(s)$

$\Delta H^o_{rxn}=-[12\times \Delta H_f(F_2)]$

(5) $2Li(s)+F_2(g)\rightarrow 2LiF(s)$

$\Delta H^o_{rxn}=-[\times \Delta H_f(F_2)]$

While the standard enthalpy of the reactions 3, 4 and 5, the $\Delta H^o{rxn}$ is not equal to $\Delta H^f{product}$ .

Hence, the reactions are,

$C(graphite)+O_2(g)\rightarrow CO_2(g)$

$BaCO_3(s)\rightarrow BaO(s)+CO_2(g)$

Answer from: AllyArce

The following reactions have $\Delta H_{{\text{rxn}}}^\circ$ equal to $\Delta H_{{\text{f}}\left( {{\text{products}}} \right)}^\circ$ :

$\boxed{{\text{C}}\left( {s,{\text{graphite}}} \right) + {{\text{O}}_2}\left( g \right) \to {\text{C}}{{\text{O}}_2}\left( g \right)}$

$\boxed{{\text{2Li}}\left( s \right) + {{\text{F}}_2}\left( g \right) \to 2{\text{LiF}}\left( s \right)}$

$\boxed{{\text{Li}}\left( s \right) + 12{{\text{F}}_2}\left( g \right) \to {\text{LiF}}\left( s \right)}$

Further explanation:

Standard enthalpy of reaction is determined by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of reactants at the standard conditions. The expression for the standard enthalpy of reaction is $\left( {\Delta H_{{\text{rxn}}}^\circ } \right)$ as follows:

$\Delta H_{{\text{rxn}}}^\circ = \sum {\text{a}}\Delta H_{{\text{f (products)}}}^\circ - \sum {\text{b}}\Delta H_{{\text{f (reactants)}}}^\circ$ ...... (1)

Here,

a is the stoichiometric coefficient of the product.

b is the stoichiometric coefficient of reactant.

$\Delta H_{{\text{f}}\left( {{\text{reactants}}} \right)}^\circ$ is the standard enthalpy for formation of reactant.

$\Delta H_{{\text{f}}\left( {{\text{products}}} \right)}^\circ$ is the standard enthalpy for formation of product.

$\Delta H_{{\text{rxn}}}^\circ$ can be equal to $\Delta H_{{\text{f}}\left( {{\text{products}}} \right)}^\circ$ only if $\Delta H_{{\text{f}}\left( {{\text{reactants}}} \right)}^\circ$ becomes equal to zero in accordance with equation (1). The value of $\Delta H_{{\text{f}}\left( {{\text{reactants}}} \right)}^\circ$ can be zero only when the reactants are present as pure elements in their natural phases.

The given reaction is,

${\text{C}}\left( {s,{\text{graphite}}} \right) + {{\text{O}}_2}\left( g \right) \to {\text{C}}{{\text{O}}_2}\left( g \right)$

On the reactant side, carbon and oxygen are present as pure elements as both are present in their respective natural phase. So $\Delta H_{{\text{f}}\left( {{\text{reactants}}} \right)}^\circ$ becomes zero and therefore $\Delta H_{{\text{rxn}}}^\circ$ is equal to $\Delta H_{{\text{f}}\left( {{\text{products}}} \right)}^\circ$ in this case.

The given reaction is,

${\text{Li}}\left( s \right) + 12{{\text{F}}_2}\left( l \right) \to {\text{LiF}}\left( s \right)$

On the reactant side, lithium is present as a pure element as its phase is same as that of its natural one. But the natural phase of fluorine is gas but in this case, it exists in liquid state. So $\Delta H_{{\text{rxn}}}^\circ$ is not equal to $\Delta H_{{\text{f}}\left( {{\text{products}}} \right)}^\circ$ in this case.

The given reaction is,

${\text{CO}}\left( g \right) + 12{{\text{O}}_2}\left( g \right) \to {\text{C}}{{\text{O}}_2}\left( g \right)$

On the reactant side, oxygen exists as a pure element as it is present in its natural phase. But CO is a compound, not an element. So $\Delta H_{{\text{rxn}}}^\circ$ is not equal to $\Delta H_{{\text{f}}\left( {{\text{products}}} \right)}^\circ$ in this case.

The given reaction is,

${\text{BaC}}{{\text{O}}_{\text{3}}}\left( s \right) \to {\text{BaO}}\left( s \right) + {\text{C}}{{\text{O}}_{\text{2}}}\left( g \right)$

On the reactant side, ${\text{BaC}}{{\text{O}}_{\text{3}}}$ is a compound, not an element. So $\Delta H_{{\text{rxn}}}^\circ$ is not equal to $\Delta H_{{\text{f}}\left( {{\text{products}}} \right)}^\circ$ in this case.

The given reaction is,

${\text{2Li}}\left( s \right) + {{\text{F}}_2}\left( g \right) \to 2{\text{LiF}}\left( s \right)$

On the reactant side, both lithium and fluorine are present as pure elements because their phases are the same as those of their natural phases. So $\Delta H_{{\text{rxn}}}^\circ$ is equal to $\Delta H_{{\text{f}}\left( {{\text{products}}} \right)}^\circ$ in this reaction.

The given reaction is,

${\text{Li}}\left( s \right) + 12{{\text{F}}_2}\left( g \right) \to {\text{LiF}}\left( s \right)$

In this case, both lithium and fluorine exist as pure elements as their phases are the same as those of their natural phases. So $\Delta H_{{\text{rxn}}}^\circ$ is equal to $\Delta H_{{\text{f}}\left( {{\text{products}}} \right)}^\circ$ in this case.

Learn more:

Calculate the enthalpy change using Hess’s Law: Find the enthalpy of decomposition of 1 mole of MgO:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Thermodynamics

Keywords: standard enthalpy of reaction, Li, F2, LiF, CO, O2, BaCO3, pure elements, compound, reactants, products.

Answer from: Quest

Flyxu7jkopvf7

Answer from: Quest

explanation:

0.07 g/cm³

1.37 g/cm³

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