Sulfur Trioxide Equilibrium

Chemistry Level 5

The oxidation of sulfur dioxide in burner gas to sulfur trioxide is important in the production of sulfuric acid. However, the conversion to sulfur dioxide decreases as temperature increases.

An equimolar mixture of S O X 2 \ce{S{ O }_{ 2 }} and O X 2 \ce{{ O }_{ 2} } was allowed to react in a rigid cylindrical vessel. The initial total pressure is 1 b a r 1 \quad bar . The reaction was carried out isothermally and the final pressure was taken at varying temperatures:

T ( ∘ C ) T (^\circ C) P f i n a l ( b a r ) { P }_{ final } (bar)
500 0.758
550 0.769
600 0.789
650 0.817
700 0.851

Determine the experimental standard entropy of the reaction in J m o l K \frac{J}{mol\quad K} . Estimate your answer to two decimal places.

Notes:

  • The reference-state partial pressure is 1 bar.

  • Assume the universal gas constant R = 8.314 J m o l K R=8.314\frac{J}{mol\quad K} .

  • Assume that the standard enthalpy and entropy of the reaction are temperature-independent.

  • Use linear regression in solving this problem.

  • Reaction must be based from one mole of sulfur dioxide.


The answer is -94.51.

Ludho Madrid by Ludho Madrid , 35, Philippines

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1 solution

Ludho Madrid
May 7, 2016

Reaction: S O 2 ( g ) + 0.5 O 2 ( g ) → S O 3 ( g ) S{ O }_{ 2\left( g \right) }+0.5{ O }_{ 2\left( g \right) }\rightarrow S{ O }_{ 3\left( g \right) }

The initial partial pressure of S O 2 S{O}_{2} and O 2 {O}_{2} are both 0.5 bar. Let x x be the extent of the reaction.

S O 2 S{O}_{2} O 2 {O}_{2} S O 3 S{O}_{3}
Initial 0.5 0.5 0.5 0.5 0 0
Change − x -x − 0.5 x -0.5x x x
Equilibrium 0.5 − x 0.5-x 0.5 − 0.5 x 0.5-0.5x x x

P f i n a l = 0.5 − x + 0.5 − 0.5 x + x = 1 − 0.5 x {P}_{final}=0.5-x+0.5-0.5x+x=1-0.5x

Find the extent of reaction per temperature:

x = 2 − 2 P f i n a l x=2-{2P}_{final}

Then, find the equilibrium constant K p {K}_{p} per temperature:

K p = x ( 0.5 − x ) 0.5 − 0.5 x {K}_{p}=\frac{x}{\left(0.5-x \right)\sqrt{0.5-0.5x}}

From the Van't Hoff Equation l n K = − Δ H R ( 1 T ) + Δ S R lnK=\frac { -\Delta H }{ R } \left( \frac { 1 }{ T } \right) +\frac { \Delta S }{ R } , graph 1 / T 1/T vs ln ⁡ K p \ln{{K}_{p}} . Perform linear regression and get the y-intercept of the regression. Note that you must convert your temperature to Kelvin.

The standard entropy of the reaction is equal to y i n t × R = − 11.367 × 8.314 J m o l K = − 94.51 J m o l K {y}_{int}\times R=-11.367\times 8.314\frac { J }{ mol\quad K } =-94.51\frac { J }{ mol\quad K } .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Thank you for posting a problem on Entropy! I love thermodynamics so much!

David Hontz - 4 years, 12 months ago

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