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CBSE Class 12-science Answered

Please provide the answer of given question
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Asked by rajeshmaheshwari80 | 16 Nov, 2018, 05:20: PM
answered-by-expert Expert Answer
 
Given:
T = 400ºC = 673.15 K
 
Let's find out Kp as we have given the equation, 
 
log subscript 10 left parenthesis Kp divided by bar right parenthesis space equals space 7.55 space minus space fraction numerator 4844 over denominator left parenthesis straight T divided by straight K right parenthesis end fraction Kp space equals space 10 to the power of open parentheses 7.55 space minus space 4844 over straight T close parentheses end exponent Kp space equals space 10 to the power of open parentheses 7.55 space minus space fraction numerator 4844 over denominator 673.15 end fraction close parentheses end exponent Kp space equals space 2.25934 space bar to the power of negative 1 end exponent Kp space equals space 2.26 space bar to the power of negative 1 end exponent
 
 
The relationship between Kp and ΔGº is given by,
 
increment straight G degree equals negative RTlnK subscript straight p superscript straight o increment straight G degree equals negative open parentheses 8.3145 space JK to the power of negative 1 end exponent mol to the power of negative 1 end exponent close parentheses space open parentheses 673.15 straight K close parentheses space ln open parentheses 2.259 bar to the power of negative 1 end exponent close parentheses increment straight G degree equals negative space 4561.885 space straight J space mol to the power of negative 1 end exponent increment straight G degree equals negative space 4.56 space KJ space mol to the power of negative 1 end exponent
 
 
The temperature dependence of the equilibrium constant is given by the equation,
 
log subscript 10 left parenthesis Kp divided by bar right parenthesis space equals space 7.55 space minus space fraction numerator 4844 over denominator left parenthesis straight T divided by straight K right parenthesis end fraction from space this space equation comma space increment straight H degree comma space by space converting space it space into space following space equation comma straight K subscript straight p superscript straight o space equals space fraction numerator increment straight H degree over denominator RT end fraction space plus space straight l
 
 
Convert log10Kp to ln Kp, 
 
 
ln space open parentheses 10 to the power of log subscript 10 Kp space end exponent close parentheses space equals space ln space Kp........................................ space log subscript straight a open parentheses straight b to the power of log subscript straight a straight x space end exponent close parentheses equals space log subscript straight a space straight x space ln space Kp equals space 2.303 space log subscript 10 Kp space 2.303 space log subscript 10 Kp space equals space Kp space equals space open parentheses negative 4844 over straight T close parentheses space 2.303 space plus space 7.55 As space we space are space solving space for space increment straight H degree comma space we space drop space the space 7.55 space term space and space multiply space by space straight R increment straight H degree space equals space space 2.303 space straight R space log subscript 10 space Kp increment straight H degree space equals space space 2.303 space open parentheses 8.3145 space straight J space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent close parentheses space open parentheses 4844 space straight K close parentheses increment straight H degree space equals space 92754.334 space Jmol to the power of negative 1 end exponent increment straight H degree space equals space 92.75 space KJmol to the power of negative 1 end exponent
 
 
The relationship between ΔG°, ΔH° and ΔS° is given by,
 
 
increment straight G space equals space increment straight H space minus space straight T increment straight S Hence comma space increment straight G degree space equals space increment straight H degree space minus space straight T increment straight S degree therefore increment straight S degree space equals space fraction numerator increment straight H degree space minus space increment straight G degree over denominator straight T end fraction therefore increment straight S degree space equals space fraction numerator 92754.334 straight J space mol to the power of negative 1 end exponent space minus open parentheses 4561.885 straight J space mol to the power of negative 1 end exponent close parentheses over denominator 673.15 straight K end fraction therefore increment straight S degree space equals space 144.568 space straight J space mol to the power of negative 1 end exponent straight K therefore increment straight S degree space equals space 145 space straight J space mol to the power of negative 1 end exponent straight K
Answered by Ramandeep | 19 Nov, 2018, 01:15: PM

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