Adiabatic Flame Temperature

Chemistry Level 5

The adiabatic flame temperature is the maximum temperature of the exiting gas after the combustion of fuel. 1 mole of LIQUID n n -octane C X 8 H X 18 \ce{{ C }_{ 8 }H_{ 18} } was burned completely in a furnace with 50% excess AIR. Assume no octane is left after combustion. Note that the nitrogen gas in air is INERT thus does not react during combustion. The liquid fuel and oxygen enter the furnace at 2 5 ∘ C 25^\circ C . Determine the adiabatic flame temperature of the exiting gas in degrees Celsius. Round your answer to four significant figures.

Notes:

  • Air composes of 79 mole % nitrogen and 21 mole % oxygen.

  • The excess air is defined as the excess amount of air with respect to the stoichiometric amount of air needed to completely combust the fuel.

Data:

  • Latent heat of vaporization of n-octane at 298.15 K: 41.53 k J / m o l 41.53 kJ/mol

  • Standard enthalpies of formation:

C 8 H 18 ( l ) = − 208.59 k J / m o l C O 2 ( g ) = − 393.51 k J / m o l H 2 O ( g ) = − 241.82 k J / m o l { C }_{ 8 }H_{ 18(l) }=-208.59\quad kJ/mol\\ C{ O }_{ 2(g) }=-393.51\quad kJ/mol\\ { H }_{ 2 }{ O }_{ (g) }=-241.82\quad kJ/mol

  • Specific heats of product gases (assumed constant):

C 8 H 18 ( g ) = 67.87 J / ( m o l K ) C O 2 ( g ) = 45.37 J / ( m o l K ) H 2 O ( g ) = 28.85 J / ( m o l K ) O 2 ( g ) = 30.25 J / ( m o l K ) N 2 ( g ) = 29.12 J / ( m o l K ) { C }_{ 8 }H_{ 18(g) }=67.87\quad J/(mol\quad K)\\ C{ O }_{ 2(g) }=45.37\quad J/(mol\quad K)\\ { H }_{ 2 }{ O }_{ (g) }=28.85\quad J/(mol\quad K)\\ { O }_{ 2(g) }=30.25\quad J/(mol\quad K)\\ { N }_{ 2(g) }=29.12\quad J/(mol\quad K)


The answer is 1796.

Ludho Madrid by Ludho Madrid , 35, Philippines

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Ludho Madrid
May 1, 2016

Reaction: C 8 H 18 ( l ) + 12.5 O 2 ( g ) → 8 C O 2 ( g ) + 9 H 2 O ( g ) { C }_{ 8 }H_{ 18(l) }+12.5{ O }_{ 2(g) }\rightarrow 8C{ O }_{ 2(g) }+9{ H }_{ 2 }{ O }_{ (g) }

Find the moles of each gas leaving the furnace.

Excess amount of O 2 { O }_{ 2 } is 12.5 × 0.5 = 6.25 m o l 12.5\times 0.5=6.25\quad mol .

Amount of inert N 2 { N }_{ 2 } leaving is ( 12.5 + 6.25 ) × 79 21 = 70.536 m o l (12.5+6.25)\times \frac { 79 }{ 21 } =70.536\quad mol .

Amount of C O 2 C{ O }_{ 2 } leaving is based on the reaction which is 8 m o l 8\quad mol .

Amount of H 2 O {H}_{2}O leaving is based on the reaction which is 9 m o l 9\quad mol .

Δ H r x n o = Δ H c o m b u s t i o n o + Δ H v a p o \Delta { H }_{ rxn }^{ o }=\Delta { H }_{ combustion }^{ o }+\Delta { H }_{ vap }^{ o } Δ H r x n o = [ 8 ( − 393.51 ) + 9 ( − 241.82 ) ] − ( − 208.59 ) + 41.53 = − 5074.34 k J m o l \Delta { H }_{ rxn }^{ o }=\left[ 8\left( -393.51 \right) +9\left( -241.82 \right) \right] -\left( -208.59 \right) +41.53=-5074.34\quad \frac { kJ }{ mol } . For 1 mole of octane, the enthalpy of the reactions is equal to − 5074340 k J -5074340\quad kJ .

Since the condition must be adiabatic, the heat from the reaction must be equal to the heat absorbed by the combustion gases. Therefore: − Δ H r x n o = q g a s e s = ∑ i r n i c i × Δ T -\Delta { H }_{ rxn }^{ o }={ q }_{ gases }=\sum _{ i }^{ r }{ { n }_{ i }{ c }_{ i } } \times \Delta T

5074340 J = [ 6.25 ( 30.25 ) + 8 ( 45.37 ) + 9 ( 28.85 ) + 70.536 ( 29.12 ) ] J K × ( T − 25 ) ℃ 5074340\quad J=\left[ 6.25\left( 30.25 \right) +8\left( 45.37 \right) +9\left( 28.85 \right) +70.536\left( 29.12 \right) \right] \quad \frac { J }{ K } \times \left( T-25 \right) ℃

Solving for T, T = 1795.93 ° C T=1795.93 °C . Rounding up to 4 significant figures, you get 1796 ° C 1796 °C .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

The solution. Is wrong The heat evolved is in kilojoules But in the 2nd last step Where you've evaluated the heat content for applying the condition for adiabaticity(heat released in reaction goes entirely into heating up the contents /mixture of gases You did not convert kilojoules to joules Hence the answer is off by a factor of 1000 Edit ☺️ . Realised there's a typo in your solution(kJ instead of J) Solved it in the same way

Suhas Sheikh - 3 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...