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Question

Heat of atomisation of $$NH_3$$ and $$N_2H_4$$ are $$x \: kcal \: mol^{-1}$$ and $$y \: kcal \: mol^{-1}$$ respectively. Average bond energy of $$N-\! \! \! -N$$ bond is:


A
3y4x3kcalmol1
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B
3yx5kcalmol1
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C
2y4x4kcalmol1
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D
None of these
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Solution

The correct option is A $$\displaystyle \frac{3y-4x}{3}\: kcal \: mol^{-1}$$
  $$NH_3\longrightarrow N(g) + 3H(g); \:        \Delta H_1=x \: kcal \: mol^{-1}$$
$$N_2H_4\longrightarrow 2N(g) + 4H(g); \:      \Delta H_2=y \: kcal \: mol^{-1}$$
                   $$ \Delta H_1=3\times e_{N-\! \! \! -H}=x$$                     ...(i)
                   $$ \Delta H_2=4\times e_{N-\! \! \! -H}+e_{N-\! \! \! -N}=y$$     ...(ii)
From eqs. (i) and (ii)
$$\displaystyle \therefore          y=4.\frac{x}{3}+e_{N-\! \! \! -N}$$
$$\displaystyle \therefore  e_{N-\! \! \! -N}=y - \frac{4x}{3}$$
              $$=\displaystyle \frac{3y-4x}{3}\: kcal \: mol^{-1}$$

Chemistry

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