Calculation Question from the student's guide to HSC Chemistry (1 Viewer)

Carl5 · Oct 31, 2011

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N<sub>2(g) </sub>+ 3H<sub>2(g) </sub>[FONT=&quot]⇌[/FONT] 2NH<sub>3(g)</sub>

At the beginning of an experiment, there were 2.1mol of nitrogen gas, and 6.9mol of hydrogen gas.
The reaction was allowed to proceed to equilibrium in a 10L container, at which point 1.2mol of N2 was remaining. What is the value of K, assuming a fixed temperature?

The first step is to note the concentration of each item individually at equilibrium. If 1.2mol of N2 is remaining, then 0.9mol must have been converted. Therefore 3 x 0.9mol of H2 must also have been converted, leaving 0.3mol of H2 (For every one mole of nitrogen gas, 3 moles of hydrogen gas are converted, as per the chemical equation). In addition, there must now be 2 x 0.9mol of NH3. We now know the concentrations of each substance. The next trick is to note that, when dealing with gases, you must account for the size of the container, as it is very rarely a simple 1L in exams. You must do this because concentration is proportional to pressure, and the size of a container influences the pressure. Once the 10L has been taken into account, K can be calculated:
K =
1.20.1 × 0.30.3
1.80.2 = 0.6310
Note that the indices have been adjusted (divided by 10) to account for the 10L container.

It says that since 0.9 mol of H2 gas has been converted. Thus, 0.9x3 mol of N2 gas has been converted. But then it says that there is now 0.3 remaining. Shouldn't the amount remaining be 6.9mol - 0.9x3 = 4.2mol?

b3kh1t · Oct 31, 2011

Carl5 said:
 · I'm rather confused.

N<sub>2(g) </sub>+ 3H<sub>2(g) </sub>[FONT=&quot]⇌[/FONT] 2NH<sub>3(g)</sub>

At the beginning of an experiment, there were 2.1mol of nitrogen gas, and 6.9mol of hydrogen gas.
The reaction was allowed to proceed to equilibrium in a 10L container, at which point 1.2mol of N2 was remaining. What is the value of K, assuming a fixed temperature?

The first step is to note the concentration of each item individually at equilibrium. If 1.2mol of N2 is remaining, then 0.9mol must have been converted. Therefore 3 x 0.9mol of H2 must also have been converted, leaving 0.3mol of H2 (For every one mole of nitrogen gas, 3 moles of hydrogen gas are converted, as per the chemical equation). In addition, there must now be 2 x 0.9mol of NH3. We now know the concentrations of each substance. The next trick is to note that, when dealing with gases, you must account for the size of the container, as it is very rarely a simple 1L in exams. You must do this because concentration is proportional to pressure, and the size of a container influences the pressure. Once the 10L has been taken into account, K can be calculated:
K =
1.20.1 × 0.30.3
1.80.2 = 0.6310
Note that the indices have been adjusted (divided by 10) to account for the 10L container.

It says that since 0.9 mol of H2 gas has been converted. Thus, 0.9x3 mol of N2 gas has been converted. But then it says that there is now 0.3 remaining. Shouldn't the amount remaining be 6.9mol - 0.9x3 = 4.2mol?

First re read the question, 0.9 mol of N2 has been converted, thus 3x0.9 mol of H2 has also been converted using the stoichiometry, but you are right when you said 6.9-2.7=4.2 mol of H2 remaining.

Also they just substituted the values into the equilibrium constant equation, that is wrong you must find the concentration first then sub in it. You can not just divide the constant you got by ten because it is essentially divided by 10^2. So find the concentration first then substitute

Calculation Question from the student's guide to HSC Chemistry (1 Viewer)

Carl5

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b3kh1t

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