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An engineer examining the oxidation of SO2 in the
Chapter 17, Problem 69P(choose chapter or problem)
An engineer examining the oxidation of \(\mathrm{SO}_{2}\) in the manufacture of sulfuric acid determines that \(K_{\mathrm{c}}=1.7 \times 10^{8} \text { at } 600 . \mathrm{K}\):
(a) At equilibrium, \(P_{\mathrm{SO}_{3}}=300\) atm and \(P_{\mathrm{O}_{2}}=100 . \mathrm{atm}\). Calculate \(P_{\mathrm{SO}_{2}}\).
(b) The engineer places a mixture of 0.0040 mol of \(\mathrm{SO}_{2}(\mathrm{g})\) and 0.0028 mol of \(\mathrm{O}_{2}(\mathrm{g})\) in a 1.0-L container and raises the temperature to 1000 K. At equilibrium, 0.0020 mol of \(\mathrm{SO}_{3}(\mathrm{g})\) is present. Calculate \(K_{\mathrm{c}}\) and \(P_{\mathrm{SO}_{2}}\) for this reaction at 1000. K.
Equation Transcription:
SO2
𝐾c = 1.7108 at 600. K
= 100
SO2(g)
O2(g)
𝐾c
Text Transcription:
SO_2
K_c=1.7 times 10^8 at 600. K
P_SO_3 = 300
P_0_2=100
P_SO_2
SO_2(g)
O_2(g)
K_c
Questions & Answers
QUESTION:
An engineer examining the oxidation of \(\mathrm{SO}_{2}\) in the manufacture of sulfuric acid determines that \(K_{\mathrm{c}}=1.7 \times 10^{8} \text { at } 600 . \mathrm{K}\):
(a) At equilibrium, \(P_{\mathrm{SO}_{3}}=300\) atm and \(P_{\mathrm{O}_{2}}=100 . \mathrm{atm}\). Calculate \(P_{\mathrm{SO}_{2}}\).
(b) The engineer places a mixture of 0.0040 mol of \(\mathrm{SO}_{2}(\mathrm{g})\) and 0.0028 mol of \(\mathrm{O}_{2}(\mathrm{g})\) in a 1.0-L container and raises the temperature to 1000 K. At equilibrium, 0.0020 mol of \(\mathrm{SO}_{3}(\mathrm{g})\) is present. Calculate \(K_{\mathrm{c}}\) and \(P_{\mathrm{SO}_{2}}\) for this reaction at 1000. K.
Equation Transcription:
SO2
𝐾c = 1.7108 at 600. K
= 100
SO2(g)
O2(g)
𝐾c
Text Transcription:
SO_2
K_c=1.7 times 10^8 at 600. K
P_SO_3 = 300
P_0_2=100
P_SO_2
SO_2(g)
O_2(g)
K_c
ANSWER:Step 1 of 4
The equilibrium constants connect the concentrations of the products and reactants. In the expression of this constant, stoichiometric integers are significant.
The reaction quotient is also expressed in a similar way and comparing its value with the equilibrium constant it indicates the direction of equilibrium shift.
(a)
Given Data:
*The equilibrium pressure of oxygen,
*The equilibrium pressure of oxygen,
*The value of 
*The temperature is 600 K.
The given reaction is:
The equilibrium constant for partial pressures for the reaction is:
The value of the equilibrium constant for partial pressures can be obtained from:
Where
R is the universal gas constant.
T is the temperature.
