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Question

For the reaction:

$$N_2 + 3H_2\rightarrow 2NH_3$$,

If $$\cfrac{d[NH_3]}{dt}= 2 \times 10^{-4} mol\ L^{-1}s^{-1}$$, the value of $$\cfrac{-d[H_2]}{dt}$$ would be:


A
1×104molL1s1
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B
3×104molL1s1
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C
4×104molL1s1
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D
6×104molL1s1
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Solution

The correct option is C $$3 \times 10^{-4} mol\; L^{-1}s^{-1}$$
          $${ N }_{ 2 }+{ 3H }_{ 2 }\longrightarrow { 2NH }_{ 3 }$$

Applying rate law.

          $$\dfrac { -1 }{ 3 } \dfrac { d\left[ { H }_{ 2 } \right]  }{ dt } =\dfrac { 1 }{ 2 } \dfrac { d\left[ { NH }_{ 3 } \right]  }{ dt } $$

Now, given that $$\dfrac { d\left[ { NH }_{ 3 } \right]  }{ dt } =2\times { 10 }^{ -4 }mol\ { L }^{ -1 }{ S }^{ -1 }$$

Using above relation,

           $$\dfrac { -d\left[ { H }_{ 2 } \right]  }{ dt } =\dfrac { 3 }{ 2 } \dfrac { d\left[ { NH }_{ 3 } \right]  }{ dt } $$

           $$\dfrac { -d\left[ { H }_{ 2 } \right]  }{ dt } =\dfrac { 3 }{ 2 } \times 2\times { 10 }^{ -4 }\ mol\ { L }^{ -1 }{ S }^{ -1 }$$

      $$\therefore $$ $$\dfrac { -d\left[ { H }_{ 2 } \right]  }{ dt } =3\times { 10 }^{ -4 }\ mol\ { L }^{ -1 }{ S }^{ -1 }$$

Hence, option B is correct.

Chemistry

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