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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

10 moles of A _ 2 , 15 moles of B _ 2 and 5 moles of AB are placed...

10molesofmoles of{A}_{2},,15molesofmoles of{B}_{2} and 5molesofmoles ofABareplacedinaare placed in a2litrevesselandallowedthecometoequilibrium.Thefinalconcentrationoflitre vessel and allowed the come to equilibrium. The final concentration ofABisis10.5M,<br/><br/>,<br /> <br />{A}_{2} \left(\text{g}\right) + {B}_{2}\left(\text{g}\right)\overset{}{⇌}}2\ttext{AB}\left(\text{g}\right)<br/><br/>Determinethevalueofequilibriumconstant(KC)br /> <br /> Determine the value of equilibrium constant \left({K}_{C}\right)for the reaction.10 moles of A _ 2 , 15 moles of B _ 2 and 5 moles of AB are placed in a 2 litre vessel and allowed the come to

Options

  1. A25.325.3
  2. B31.531.5
  3. C36.336.3
  4. D40.540.5

Correct answer

B. 31.531.5

Step-by-step solution

10 moles of A _ 2 , 15 moles of B _ 2 and 5 moles of AB are placed in a 2 litre vessel and allowed the come to

5+2x2=10.5;x=8\dfrac{5 + 2x}{2} = 10.5; \text{x} = 8

KC=[AB]2[A2][B2]=(5+2x2)2(10−x2)(15−x2)=(21)22×7=31.5{K}_{C} = \dfrac{{[ AB ]}^{2}}{\left[{A}_{2}\right]\left[{B}_{2}\right]} = \dfrac{{\left(\dfrac{5 + 2x}{2}\right)}^{2}}{\left(\dfrac{10 - x}{2}\right)\left(\dfrac{15 - x}{2}\right)} = \dfrac{{( 21 )}^{2}}{2 \times 7} = 31.5
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