State second law of thermodynamics in terms of entropy. from Ch

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 Multiple Choice QuestionsShort Answer Type

281.

For the equilibrium
PCl subscript 5 space space rightwards harpoon over leftwards harpoon space space space space PCl subscript 3 left parenthesis straight g right parenthesis space plus space Cl subscript 2 left parenthesis straight g right parenthesis space at space 298 space straight K comma straight K space equals space 1.8 space cross times space 10 to the power of negative 7 end exponent
What is the ∆G0 for the reaction?

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282. The equilibrum constant for a reaction is 10. What will be the value of ∆G0 ? R = 8.314 JK–1 mol–1, T = 300 K. 
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283.

Calculate increment subscript straight r straight G to the power of 0 for conversion of oxygen to ozone, 3 over 2 straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space space space straight O subscript 2 left parenthesis straight g right parenthesis space at space 298 space straight K comma If straight K subscript straight P for this conversation is 2.47 space cross times space 10 to the power of negative 29 end exponent.

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 Multiple Choice QuestionsLong Answer Type

284.

Calculate the equilibrium constant for the reaction at 400 K:

space space 2 NOCl left parenthesis straight g right parenthesis space space space rightwards harpoon over leftwards harpoon space space space 2 NO left parenthesis straight g right parenthesis space plus space Cl subscript 2 left parenthesis straight g right parenthesis
increment subscript straight r straight H to the power of 0 space equals space 77.2 space KJ space mol to the power of negative 1 end exponent space and
increment subscript straight r straight S to the power of 0 space equals space 122 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent space 400 space straight K.
            
                                
       
       

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 Multiple Choice QuestionsShort Answer Type

285.

Find out the value of equilibrium constant for the following reaction at 298 K:

2 NH subscript 3 left parenthesis straight g right parenthesis space plus space CO subscript 2 left parenthesis straight g right parenthesis space space space rightwards harpoon over leftwards harpoon space space NH subscript 2 CONH subscript 2 left parenthesis aq right parenthesis space plus space straight H subscript 2 straight O left parenthesis l right parenthesis

   Standard Gibbs energy change,  increment straight G degree at the given temperature, is -13.6 kJ mol-1.

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286.

The equilibrium constant at 25 degree straight C for the process  <pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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Calculate the value of increment subscript straight r straight G to the power of 0 space at space 25 degree straight C space left parenthesis straight R space equals space 8.314 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent right parenthesis. In which direction is the reaction spontaneous when reactants and products are under standard conditions?

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 Multiple Choice QuestionsLong Answer Type

287. At 60°C, dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere. 
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 Multiple Choice QuestionsShort Answer Type

288.

Using the data given below, calculate the value of equilibrium constant for the reaction

stack 3 HC identical to CH left parenthesis straight g right parenthesis with Acetylene below space space rightwards harpoon over leftwards harpoon space space straight C subscript 6 straight H subscript 6 left parenthesis straight g right parenthesis space at space 298 space straight K

assuming ideal behaviour.

increment subscript straight r straight G to the power of 0 space left parenthesis HC space identical to space CH right parenthesis space equals space 2.09 space cross times space 10 to the power of 5 space straight J space mol to the power of negative 1 end exponent comma space
increment subscript straight r straight G to the power of 0 left parenthesis straight C subscript 6 straight H subscript 6 right parenthesis space equals space 1.24 space cross times space 10 to the power of 5 space straight J space mol to the power of negative 1 end exponent space and
straight R space equals space 8.314 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent                       

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289.

State second law of thermodynamics in different ways.

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290.

State second law of thermodynamics in terms of entropy.


All spontaneous processes are accompanied by a net increase of entropy, i.e. for all spontaneous processes, the total entropy change (sum of the entropy changes of the system and the surroundings) is positive.
Since all naturally occurring processes are spontaneous and are accompanied by a net increase of entropy, therefore the second law of thermodynamics may also be stated as: “The entropy of the inverse is continuously increasing and tends to be maximum.

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