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3 years ago

13

In the Haber process, nitrogen gas is combined with hydrogen (from natural gas) to form ammonia. If ammonia is formed at 0.345 M

/s, how quickly is the nitrogen gas disappearing

1 answer:

algol133 years ago

6 0

Answer:

$r_{N_2}=-0.1725M/s$

Explanation:

Hello,

In this case, by means of the law of mass action, we firstly write the described chemical reaction:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

Thus, as ammonia is being formed at 0.345 M/s, nitrogen will be disappearing at (consider law of mass action):

$r_{NH_3}=0.345M/s\\\\\frac{1}{-1} r_{N_2}=\frac{1}{-3}r_{H_2}=\frac{1}{2} r_{NH_3}\\\\r_{N_2}=-\frac{1}{2} r_{NH_3}=-\frac{1}{2} *0.345M/s\\\\r_{N_2}=-0.1725M/s$

Best regards.

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We need to determine the empirical formula in its simplest form, where hydrogen ( $H$ ) is scaled up to a mole, since it has the molar mass, and both carbon ( $C$ ) and oxygen ( $O$ ) are also scaled up in the same magnitude. The empirical formula is of the form:

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$M_{H}$ - Molar mass of hydrogen, in grams per mole.

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$n_{C} = \frac{r\cdot m_{C}}{M_{C}}$ (2)

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Hence, the empirical formula of the compound is $C_{0.504}HO_{1.008}$ .

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