1) The forward reaction is N2 (g) + O2 (g) → 2NO

(that reaction requires special contitions because at normal pressures and temperatures N2 and O2 do not react to form another compound.

2) The equiblibrium equation is

N2 (g) + O2 (g) ⇄ 2NO

3) Then, the reverse reaction is

2NO → N2(g) + O2(g)

Answer: 2NO → N2(g) + O2(g)

4 0

3 years ago

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

grin007 [14]

Answer:

$n_{C2H_5OH}^{eq}=14.234mol$

Explanation:

Hello,

In this case, the reaction is:

$C_2H_4+H_2O\rightleftharpoons CH_3CH_2OH$

Thus, the law of mass action turns out:

$Kc=\frac{[CH_3CH_2OH]_{eq}}{[H_2O]_{eq}[CH_2CH_2]_{eq}}$

Thus, since at the beginning there are 29 moles of ethylene and once the equilibrium is reached, there are 16 moles of ethylene, the change $x$ result:

$[CH_2CH_2]_{eq}=29mol-x=16mol\\x=29-16=13mol$

In such a way, the equilibrium constant is then:

$Kc=\frac{\frac{x}{V} }{\frac{16mol}{V}* \frac{3mol}{V}} =\frac{\frac{13mol}{75.0L} }{\frac{16mol}{75.0L}* \frac{3mol}{75.0L}} =20.31$

Thereby, the initial moles for the second equilibrium are modified as shown on the denominator in the modified law of mass action by considering the added 15 moles of ethylene:

$Kc=\frac{\frac{13+x_2}{V} }{\frac{16+15-x_2}{V}* \frac{3-x_2}{V}} =20.31$

Thus, the second change, $x_2$ finally result (solving by solver or quadratic equation):

$x_2=1.234mol$

Finally, such second change equals the moles of ethanol after equilibrium based on the stoichiometry:

$n_{C2H_5OH}^{eq}=x+x_2=13mol+1.234mol\\n_{C2H_5OH}^{eq}=14.234mol$

Best regards.

5 0

3 years ago

Consider just the first sentence of this multistep word problem. Sentence 1: The sum of the ages of a brother and sister is 26.

kiruha [24]

Answer:

a. b + s = 26

b. b = 26 — s

Explanation:

5 0

3 years ago

CAN SOMEONE HELP ME PLZ AND THANKS WILL MARK U AS BRAINLIEST

jekas [21]

Explanation:

2. $2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$

First, we need to find the number of moles of $CO_2$ at 300K and 1.5 atm using the ideal gas law:

$n= \dfrac{PV}{RT}= \dfrac{(1.5\:\text {atm})(33\:L)}{(0.082\:\text{L-atm/mol-K})(300K)}$

$=2.0\:\text{mol}\:CO_2$

Now use the molar ratios to find the number of moles of ethane to produce this much $CO_2$ .

$2.0\:\text{mol}\:CO_2 \times \left(\dfrac{2\:\text{mol}\:C_2H_6}{4\:\text{mol}\:CO_2}\right)$

$=1.0\:\text{mol}\:C_2H_6$

Finally, convert this amount to grams using its molar mass:

$1.0\:\text {mol}\:C_2H_6 \times \left(\dfrac{30.07\:\text g\:C_2H_6}{1\:\text{mol}\:C_2H_6} \right)$

$=30.1\:g\:C_2H_6$

3. $3Zn + 2H_3PO_4 \rightarrow 3H_2 + Zn_3(PO_4)_2$

Convert 75 g Zn into moles:

$75\:\text g\:Zn \times \left(\dfrac{65.38\:\text g\:Zn}{1\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:Zn$

Then use the molar ratios to find the amount of H2 produced.

$1.1\:\text{mol}\:Zn \times \left(\dfrac{3\:\text{mol}\:H_2}{3\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:H_2$

Now use the ideal gas law $PV=nRT$ to find the volume of H2 produced at 23°C and 4 atm:

$V= \dfrac{nRT}{P}= \dfrac{(1.1\:\text{mol}\:H_2)(0.082\:\text{L-atm/mol-K})(296K)}{4\:\text{atm}}$

$=8.9\:\text L\:H_2$

8 0

2 years ago