`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`
`color{green}("Definition" )` The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called Standard Molar Enthalpy of Formation.
● Its symbol is `color{purple}(Δ_fH^⊖)`, where the subscript `‘ f ’` indicates that one mole of the compound in question has been formed in its standard state from its elements in their most stable states of aggregation.
● The reference state of an element is its most stable state of aggregation at `25°C` and `1` bar pressure.
`color{red}("Example ")` The reference state of dihydrogen is `H_2` gas and those of dioxygen, carbon and sulphur are `O_2` gas, `color{purple}(C_text(graphite))` and `color{purple}(S_text(rhombic))` respectively.
● Some reactions with standard molar enthalpies of formation are given below.
`color{purple}(H_2(g) +1/2 O_2 (g) → H_2O(l)) ; `
`color{purple}(Delta_f H^⊖ = -285 kJ mol^(-1))`
`color{purple}(C_text(graphite, s) +2H_2 (g)→ CH_4(g)); `
`color{purple}(Δ_f H^⊖ = −74.8kJmol^(-1))`
`color{purple}(2C_text{(graphite,s)} +3H_2(g) +1/2 O_2(g) → C_2H_5OH (l)) ; `
`color{purple}(Delta_f H^⊖ = -277.7 kJ mol^(-1))`
`color{red}("Note ")` (i) Standard molar enthalpy of formation, `color{purple}(Δ_fH^⊖)`, is just a special case of `Δ_rH^⊖`, where one mole of a compound is formed from its constituent elements, as in the above three equations, where `1` mol of each, water, methane and ethanol is formed.
(ii) In contrast, the enthalpy change for an exothermic reaction :
`color{purple}(CaO(s) + CO_2 (g) → CaCO_3(s)) ; `
`color{purple}(Delta_fH^⊖ = - 178 kJ mol^(-1))`
is not an enthalpy of formation of calcium carbonate, since calcium carbonate has been formed from other compounds, and not from its constituent elements.
(iii) Also, for the reaction given below, enthalpy change is not standard enthalpy of formation, `color{purple}(Δ_fH^⊖)` for `color{purple}(HBr(g))`.
`color{purple}(H_2 (g) +Br_2 (l) →2HBr (g)) ; `
`color{purple}(Delta_f H^⊖ = -72.8 kJ mol^(-1))`
Here two moles, instead of one mole of the product is formed from the elements, i.e.,
`color{purple}(Delta_r H^⊖ = 2 Delta_f H^⊖)`
● Therefore, by dividing all coefficients in the balanced equation by 2, expression for enthalpy of formation of `HBr (g)` is written as
`color{purple}(1/2 H_2 (g) +1/2 Br_2(l) → HBr (g)) ; `
`color{purple}(Delta_fH^⊖ = -36.4 kJ mol^(-1))`
`=>` Standard enthalpies of formation of some common substances are given in Table 6.2.
`=>` By convention, standard enthalpy for formation, `color{purple}(Δ_fH^⊖),` of an element in reference state, i.e., its most stable state of aggregation is taken as zero.
`=>` Suppose, you are a chemical engineer and want to know how much heat is required to decompose calcium carbonate to lime and carbon dioxide, with all the substances in their standard state.
`color{purple}(CaCO_3 (s)→CaO(s) +CO_2 (g) ; Delta_r H^⊖ = ?)`
● Here, we can make use of standard enthalpy of formation and calculate the enthalpy change for the reaction. The following general equation can be used for the enthalpy change calculation.
`color{purple}(Delta_r H^⊖ = underset(i)Sigma a_i Delta_f H^⊖ ("products") - underset(i)Sigma b_i Delta_f H^⊖) ("reactants"))` .........(6.15)
where `color{purple}(a)` and `color{purple}(b)` represent the coefficients of the products and reactants in the balanced equation.
● Let us apply the above equation for decomposition of calcium carbonate. Here, coefficients ‘a’ and ‘b’ are `1` each. Therefore,
`color{purple}(Delta_r H^⊖ = Delta_f H^⊖ [CaO(s) ]+Delta_gH^⊖ [CO_2(g)] - Delta_f H^⊖ [CaCO_3(s)])`
` color{purple}(=1(-635.1 kJ mol^(-1) ) +1(-393.5kJ mol^(-1) )-1 (-1206.9 kJ mol^(-1) ))`
`color{purple}(= 178.3 kJ mol^(-1))`
● Thus, the decomposition of `color{purple}(CaCO_3)` (s) is an endothermic process and you have to heat it for getting the desired products.
`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`
`color{green}("Definition" )` The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called Standard Molar Enthalpy of Formation.
● Its symbol is `color{purple}(Δ_fH^⊖)`, where the subscript `‘ f ’` indicates that one mole of the compound in question has been formed in its standard state from its elements in their most stable states of aggregation.
● The reference state of an element is its most stable state of aggregation at `25°C` and `1` bar pressure.
`color{red}("Example ")` The reference state of dihydrogen is `H_2` gas and those of dioxygen, carbon and sulphur are `O_2` gas, `color{purple}(C_text(graphite))` and `color{purple}(S_text(rhombic))` respectively.
● Some reactions with standard molar enthalpies of formation are given below.
`color{purple}(H_2(g) +1/2 O_2 (g) → H_2O(l)) ; `
`color{purple}(Delta_f H^⊖ = -285 kJ mol^(-1))`
`color{purple}(C_text(graphite, s) +2H_2 (g)→ CH_4(g)); `
`color{purple}(Δ_f H^⊖ = −74.8kJmol^(-1))`
`color{purple}(2C_text{(graphite,s)} +3H_2(g) +1/2 O_2(g) → C_2H_5OH (l)) ; `
`color{purple}(Delta_f H^⊖ = -277.7 kJ mol^(-1))`
`color{red}("Note ")` (i) Standard molar enthalpy of formation, `color{purple}(Δ_fH^⊖)`, is just a special case of `Δ_rH^⊖`, where one mole of a compound is formed from its constituent elements, as in the above three equations, where `1` mol of each, water, methane and ethanol is formed.
(ii) In contrast, the enthalpy change for an exothermic reaction :
`color{purple}(CaO(s) + CO_2 (g) → CaCO_3(s)) ; `
`color{purple}(Delta_fH^⊖ = - 178 kJ mol^(-1))`
is not an enthalpy of formation of calcium carbonate, since calcium carbonate has been formed from other compounds, and not from its constituent elements.
(iii) Also, for the reaction given below, enthalpy change is not standard enthalpy of formation, `color{purple}(Δ_fH^⊖)` for `color{purple}(HBr(g))`.
`color{purple}(H_2 (g) +Br_2 (l) →2HBr (g)) ; `
`color{purple}(Delta_f H^⊖ = -72.8 kJ mol^(-1))`
Here two moles, instead of one mole of the product is formed from the elements, i.e.,
`color{purple}(Delta_r H^⊖ = 2 Delta_f H^⊖)`
● Therefore, by dividing all coefficients in the balanced equation by 2, expression for enthalpy of formation of `HBr (g)` is written as
`color{purple}(1/2 H_2 (g) +1/2 Br_2(l) → HBr (g)) ; `
`color{purple}(Delta_fH^⊖ = -36.4 kJ mol^(-1))`
`=>` Standard enthalpies of formation of some common substances are given in Table 6.2.
`=>` By convention, standard enthalpy for formation, `color{purple}(Δ_fH^⊖),` of an element in reference state, i.e., its most stable state of aggregation is taken as zero.
`=>` Suppose, you are a chemical engineer and want to know how much heat is required to decompose calcium carbonate to lime and carbon dioxide, with all the substances in their standard state.
`color{purple}(CaCO_3 (s)→CaO(s) +CO_2 (g) ; Delta_r H^⊖ = ?)`
● Here, we can make use of standard enthalpy of formation and calculate the enthalpy change for the reaction. The following general equation can be used for the enthalpy change calculation.
`color{purple}(Delta_r H^⊖ = underset(i)Sigma a_i Delta_f H^⊖ ("products") - underset(i)Sigma b_i Delta_f H^⊖) ("reactants"))` .........(6.15)
where `color{purple}(a)` and `color{purple}(b)` represent the coefficients of the products and reactants in the balanced equation.
● Let us apply the above equation for decomposition of calcium carbonate. Here, coefficients ‘a’ and ‘b’ are `1` each. Therefore,
`color{purple}(Delta_r H^⊖ = Delta_f H^⊖ [CaO(s) ]+Delta_gH^⊖ [CO_2(g)] - Delta_f H^⊖ [CaCO_3(s)])`
` color{purple}(=1(-635.1 kJ mol^(-1) ) +1(-393.5kJ mol^(-1) )-1 (-1206.9 kJ mol^(-1) ))`
`color{purple}(= 178.3 kJ mol^(-1))`
● Thus, the decomposition of `color{purple}(CaCO_3)` (s) is an endothermic process and you have to heat it for getting the desired products.