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CBSE Class 12-science Answered

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Asked by Balbir | 02 Oct, 2019, 12:24: AM
answered-by-expert Expert Answer
Firstly, We have to calculate change in Gibbs free energy.
 
C r to the power of plus 2 end exponent space plus 3 e to the power of minus space rightwards arrow C r subscript left parenthesis s right parenthesis end subscript space space space space....... left parenthesis 1 right parenthesis space space space increment G to the power of o subscript 1 equals negative 3 F E to the power of o subscript bevelled fraction numerator C r to the power of plus 3 end exponent over denominator C r end fraction end subscript C r to the power of plus 3 end exponent space plus e to the power of negative space end exponent rightwards arrow C r to the power of plus 2 end exponent space space space space space........ left parenthesis 2 right parenthesis space space space space space increment G to the power of o subscript 2 equals negative F E to the power of o subscript bevelled fraction numerator C r to the power of plus 3 end exponent over denominator C r to the power of plus 2 end exponent end fraction end subscript left parenthesis 1 right parenthesis minus left parenthesis 2 right parenthesis C r to the power of plus 2 end exponent space plus space 2 e to the power of minus rightwards arrow C r subscript left parenthesis s right parenthesis end subscript increment G to the power of o equals negative 3 cross times 0.5 F minus 0.41 F space space space space space space space equals negative 1.91 F E to the power of o subscript C r to the power of plus 2 end exponent divided by C r end subscript equals fraction numerator 1.91 over denominator 2 end fraction equals 0.955 space V E to the power of o subscript c e l l end subscript equals E to the power of o subscript C r to the power of plus 2 end exponent divided by C r end subscript minus E to the power of o subscript C r to the power of plus 3 end exponent divided by C r to the power of plus 2 end exponent end subscript space space space space space space space space equals 0.955 plus 0.41 space space space space space space space space equals 1.365 thin space V increment G to the power of o equals negative n F E to the power of o subscript c e l l end subscript space space space space space space space space equals negative 2 cross times 96500 cross times 1.365 space space space space space space space space equals negative 263.44 space K J minus 263.44 equals increment H plus 298 open parentheses fraction numerator negative 270.56 plus 263.44 over denominator 10 end fraction close parentheses increment H equals negative 53.05 space K J increment G equals increment H minus T increment S space space space space space space equals fraction numerator negative 53.05 plus 263.44 over denominator 298 end fraction increment S equals 0.706 space K J K to the power of negative 1 end exponent space space space
Answered by Ravi | 07 Oct, 2019, 03:29: PM
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