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jasenka [17]
3 years ago
14

For the oxidation of ammonia

Chemistry
1 answer:
OverLord2011 [107]3 years ago
5 0

The answer is 0.405 M/s

- (1/3) d[O2]/dt = 1/2 d[N2]/dt

- d[O2]/dt = 3/2 d[N2]/dt

- d[O2]/dt = 3/2 × 0.27

- d[O2]/dt = 0.405 mol L^(-1) s^(-1)

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Calculate the change in the standard entropy of the system, delta s degree for the synthesis of ammonia from N_2(g) and H_2(g) a
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Answer:

\Delta S^{0} for the reaction is -198.762 J/K

Explanation:

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Standard change in entropy for the system (\Delta S^{0}) is given by-

\Delta S^{0}=[2moles\times S^{0}(NH_{3})_{g}]-[1mole\times S^{0}(N_{2})_{g}]-[3\times S^{0}(H_{2})_{g}]

where S^{0} represents standard entropy.

Here S^{0}(NH_{3})_{g}=192.45J/(K.mol), S^{0}(N_{2})_{g}=191.61J/(K.mol) and S^{0}(H_{2})_{g}=130.684J/(K.mol)

So, \Delta S^{0}=[2\times 192.45]-[1\times 191.61]-[3\times 130.684]J/K=-198.762J/K

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