Predict the effect of an eightfold pressure increase on the equilibriunm composition of the reaction 3 Nalg) + H21g) =2 NaHlg)

2 answers:

6 0

According to Le cha telier's principle when we increase the pressure of a equilibrium reaction the reaction shift to the side where few moles of gas present.

Your reaction (Notice - I guess the reaction you written is wrong, but still I'm solving with your given prediction)

3NaI (g) + H2 (g) = 2NaHI (g)

Where access of sodium iodide is reacting with Hydrogen gas to form NaHI molecule.

Number of moles of gas on reaction side - 3+2 = 5 moles

Number of moles of gas on product side - 2 moles

<em>Conclusion- the reaction will shift to the right of the reaction</em>

Assoli18 [71]3 years ago

4 0

According to Le cha telier's principle when we increase the pressure of a equilibrium reaction the reaction shift to the side where few moles of gas present.

Your reaction (Notice - I guess the reaction you written is wrong, but still I'm solving with your given prediction)

3NaI (g) + H2 (g) = 2NaHI (g)

Where access of sodium iodide is reacting with Hydrogen gas to form NaHI molecule.

Number of moles of gas on reaction side - 3+2 = 5 moles

Number of moles of gas on product side - 2 moles

Conclusion- the reaction will shift to the right of the reaction

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$100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{1molO}{1molH20}*\frac{16gr}{1molO}*\frac{1h}{3600s}=11,2\frac{gO}{s}$

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$100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{2molH}{1molH20}*\frac{1gr}{1molH}*\frac{1h}{3600s}=1,4\frac{gH}{s}$

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saveliy_v [14]

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