Equivalence Ratio, Stoichiometric Ratio

4. Equivalence Ratio, Stoichiometric Ratio

The equivalence ratio φ = (F:A)/(F:A) stoich = (A:F) stoich /(A:F) = (O 2 :F) stoich /(O 2 :F)

(6) As an example, if φ = 0.5 for methane–air combustion, this implies that the excess is air sup-

plied for every kmole of fuel that is burned. In general, for methane–air combustion CH 4 + (2/ φ)(O 2 + 3.76 N 2 ) → CO 2 + 2H 2 O + 2((1/ φ)–1) O 2 + (2/ φ)×3.76N 2 , or (7)

Another term used to represent the fuel–air mixture composition is the stoichiometric ratio

SR = (actual air supplied)/(stoichiometric air demand of fuel) = 1/ φ.

For instance, if φ = 0.5, then SR = 2, i.e., the air supplied is two times as large as the stoichiometric or theoretical air demand of the fuel.

a. Example 1 Consider the metabolism of glucose in the human body. As we breathe in air, we transfer oxygen from our lungs into our bloodstream. That oxygen is transported to the cells of our tissues where it oxidizes glucose. Write down the stoichiometric reac-

tion for the consumption of glucose (s) C 6 H 12 O 6 by pure oxygen and by air. Deter- tion for the consumption of glucose (s) C 6 H 12 O 6 by pure oxygen and by air. Deter-

is consumed per minute? Solution The stoichiometric or theoretical reaction equation for this case is

(A) Applying an atom balance,

C 6 H 12 O 6 +aO 2 → b CO 2 +cH 2 O

Carbon C: 6 = b, (B) Hydrogen H: 12 = 2 c, i.e., c = 6, and

(C) Oxygen O: 6 + 2a = 2 b + c, i.e., a = 6.

(D) Therefore, the stoichiometric relation assumes the form

(E) The corresponding stoichiometric or theoretical reaction equation in the case of air is

C 6 H 12 O 6 +6O 2 → 6 CO 2 +6H 2 O.

C 6 H 12 O 6 + 6 (O 2 + 3.76 N 2 ) → 6 CO 2 +6H 2 O + 22.56 N 2 . (F) When excess air is supplied, 400% excess air = (supplied air–stoichiometric air) ×100/stoichiometric air

= (supplied air – (6 + 22.56)) × 100 ÷ (6+22.56), i.e., the supplied air = 142.8 kmole of air/kmole of glucose.

(G) Therefore, the excess air as a percentage of theoretical air equals (A:F/(A:F) stoich ) × 100 = 142.8÷28.56 = 500%.

(H) The equivalence ratio

φ = ((A:F) stoich /(A:F)) = 28.56 ÷142.8 = 0.2. (I) The actual reaction equation with 400% excess air or five times the theoretical air is

C 6 H 12 O 6 +5 ×6(O 2 + 3.76 N 2 ) → 6CO 2 +6H 2 O+aO 2 +bN 2 . (J) Applying the atom balance for Oxygen O: 5 ×6×2 = 6×2 + 6×1 + 2a, i.e., a = 24, and

(K) Nitrogen N: b = 30 ×2×3.76÷2 = 112.8, i.e.,

(L)

C 6 H 12 O 6 +5 ×6 (O 2 + 3.76 N 2 ) → 6CO 2 +6H 2 O + 24 O 2 + 112.8 N 2 . (M) The air consumption is specified as 360 L hr ––1 or 0.0147 kmole hr –1 . Therefore, the

glucose consumption per hour is ω = 0.0147÷142.8 = 0.0001029 kmole hr –1 .

(N)

The mass of one kmole of glucose is 180.2 kg, so that the glucose consumption per hour in terms of mass is

m = 00.0001029 kmole hr –1 × 180.2 kg kmole –1 = 18.6 g hr –1 or 0.31 g min –1 . (O) Remarks

The ratio of CO 2 to O 2 is called respiratory quotient (RQ) in the medical literature. For glucose, the stoichiometric reaction indicates that RQ =1. Fats (e.g., palmitic acid) are also used by the body for metabolism. The stoichiometric reaction for fat is given as

C 15 H 31 COOH + 23 O 2 → 16CO 2 + 16 H 2 O.

The RQ is given as 0.7. Since older persons have problems with excreting CO 2 , then fats are preferable compared to glucose due to lower RQ values.