Chemistry-
General
Easy

Question

The reaction S subscript 2 end subscript O subscript 8 end subscript superscript 2 minus end superscript plus 3 I to the power of minus end exponent rightwards arrow 2 S O subscript 4 end subscript superscript 2 minus end superscript plus I subscript 3 end subscript superscript minus end superscriptis of first order both with respect to the persulphate and iodide ions. Taking the initial concentration as ' a ' and 'b' respectively and taking x as the concentration of the triodide at time t a differential rate equation can be written. Two suggested mechanisms for the reaction are

The general differential equation for the above reaction is

  1. fraction numerator d x over denominator d t end fraction equals k left square bracket a minus x right square bracket left square bracket b minus 3 x right square bracket    
  2. fraction numerator d x over denominator d t end fraction equals negative k left square bracket a minus x right square bracket left square bracket b minus 3 x right square bracket    
  3. fraction numerator d x over denominator d t end fraction equals k left square bracket a minus x right square bracket left square bracket b minus x right square bracket    
  4. fraction numerator d x over denominator d t end fraction equals negative k left square bracket a minus x right square bracket left square bracket b minus x right square bracket    

The correct answer is: fraction numerator d x over denominator d t end fraction equals k left square bracket a minus x right square bracket left square bracket b minus 3 x right square bracket

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Figure shows a graph in l o g subscript 10 end subscript invisible function application K v s fraction numerator 1 over denominator text end text T end fraction where K is rate constant and T is temperature. The straight line BC has slope, T a n invisible function application theta equals negative fraction numerator 1 over denominator 2.303 end fraction and an intercept of 5 on y-axis. Thus E subscript a end subscript, the energy of activation is:

Figure shows a graph in l o g subscript 10 end subscript invisible function application K v s fraction numerator 1 over denominator text end text T end fraction where K is rate constant and T is temperature. The straight line BC has slope, T a n invisible function application theta equals negative fraction numerator 1 over denominator 2.303 end fraction and an intercept of 5 on y-axis. Thus E subscript a end subscript, the energy of activation is:

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For an exothermic chemical process occurring in two steps as :
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The progress of the reaction can be best described by :

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The heats of combustion of carbon, hydrogen and acetylene are -394K.J, -286K.J and -1301 K.J respectively. Calculate heat of formation of C subscript 2 end subscript H subscript 2 end subscript

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For the adsorption of solution on a solid surface fraction numerator x over denominator m end fraction equals k c to the power of 1 divided by n end exponent Adsorption isotherm of l o g invisible function application open parentheses fraction numerator x over denominator m end fraction close parentheses l o g invisible function application C was found of the type (Fig) This is when

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chemistry-General
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maths-General
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