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This is the second of our many quizzes on Kinetics. Click on this to access more such competitive quizzes on Physical Chemistry.

Kinetics Quiz 2

chemistry

 The progress of a reaction $A\rightleftharpoons nB$ is represented in the graph below<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5bbc6cbf-2cd6-4496-baf9-b70b3b3a5e703818944699428599633.png"/><br>What will be the value of the equilibrium constant?	 

 n can be calculated by taking the ratio of the increase in the concentration of product and decrease in the concentration of the reactant in a given time.

In a time frame of 10 hours (2 to 12), the concentration of A decreases by 0.6M (1.0 to 0.4), and the concentration of B increases by 0.8M (0.4 to 1.2). <br>$ \\ \text{So, n = }\dfrac{0.8}{0.6}=\dfrac{4}{3}.\\ \text{Hence equilibrium constant will be }\\ K=\dfrac{{{(1.2)}^{\dfrac{4}{3}}}}{(0.4)}\cong 3.2M$ 

 Which of the following is the correct rate expression for a reaction whose order is 3.

 The sum of powers of the concentrations of the reactants in the rate law expression is equal to the order of the reaction.

Since for all the options, the sum of powers of the concentrations of the reactants is equal to 3, hence the order of the reaction is 3 in all the options.

 The rate constants of a reaction at 200K and 100K are in the ratio of 4:3. The value of activation energy will be

 The Arrhenius equation $k=A{{e}^{\dfrac{-{{E}_{a}}}{RT}}}$ can be used to derive the relation between activation energy and the rate constant at different temperatures. 

 $ \log \dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}\times {{T}_{2}}} \right] \\ \log \dfrac{4}{3}=\dfrac{{{E}_{a}}}{2.303\times 8.314}\left[ \dfrac{200-100}{200\times 100} \right] \\ {{E}_{a}}=\log \dfrac{4}{3}\times 2.303\times 8.314\times 200 \\ {{E}_{a}}\cong 478.44J$

 The ratio of the initial concentration of the reactant and the concentration of the reactant after 37s in a first-order reaction is 12:5. The rate constant of the reaction will be approximately

 The rate of the reaction in a first-order reaction is only dependent on one reactant.

 $ k=\dfrac{2.303}{t}\log \dfrac{{{\left[ R \right]}_{0}}}{\left[ R \right]} \\ k=\dfrac{2.303}{37}\log \dfrac{{{\left[ 12 \right]}_{0}}}{\left[ 5 \right]} \\ k=2.37\times 1{{0}^{-2}}{{s}^{-1}}$

 Which of the following reactions is bimolecular?

  The number of reacting molecules defines the molecularity of a reaction.

 Since there are 2 reactant molecules in reaction (a), it is a bimolecular reaction. Reaction (b) is unimolecular and reaction (c) is trimolecular.