
Consider the following reaction equilibrium: . Initially, \[1\] mole of \[{N_2}\] and \[3\] moles of \[{H_2}\] are taken in a \[2\] litre flask.
At equilibrium state, if the number of moles of \[{N_2}\] is \[0.6\], then the total number of moles of all gases present in the flask are:
A.\[0.8\]
B.\[1.6\]
C.\[2.8\]
D.\[3.2\]
Answer
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Hint: According to the given equation, \[1\] mole of nitrogen and \[3\] moles of hydrogen gives \[2\] moles of ammonia. Consider the non-equilibrium gases to have an arbitrary number of moles. Use it to get the equation of all the gases which are present in the flask. This will give us the value of an arbitrary number of moles which will be further used to solve the equation of gases. Just add all the equations of gases to get the value of the total number of moles of all gases present in the flask.
Complete answer:
In the equilibrium equation: , initially, \[1\] mole of \[{N_2}\], \[3\] moles of \[{H_2}\] and \[0\] moles of ammonia are present. Let the number of moles of gases not present in equilibrium be \[x\]. Then,
Number of moles of \[{N_2} = 1 - x\]
Number of moles of \[{H_2} = 3 - 3x\]
Number of moles of \[N{H_3} = 2x\]
Also, in the equilibrium state, \[0.6\] moles of \[{N_2}\] is present. Hence,
Number of moles of \[{N_2} = 1 - x = 0.6\]
\[x = 0.4\]
Number of moles of \[{H_2} = 3 - 3 \times 0.4 = 3 - 1.2 = 1.8\]
Number of moles of \[N{H_3} = 2x = 2 \times 0.4 = 0.8\]
Therefore, \[0.4\] moles of nitrogen and \[1.8\] moles of hydrogen react to form \[0.8\] moles of ammonia.
Total number of moles of gases present in the flask is \[ = 1 - x + 3 - 3x + 2x\]
\[ = 4 - 2x\]
On putting the value of \[x\], we get,
\[ = 4 - 2 \times 0.4\]
\[ = 4 - 0.8\]
\[ = 3.2\]
Hence, the total number of moles of all the gases present in the flask is \[3.2\].
Hence, Option D is correct.
Note:
When solving the individual equation of gases, students might replace the arbitrary value by \[0.6\]. This will give the correct result for the first equation but it will give an incorrect result for the rest of the equation. And finally adding all the equations will result in an incorrect result. The final addition is also necessary as we are asked for the moles of all the gases present in the flask.
Complete answer:
In the equilibrium equation: , initially, \[1\] mole of \[{N_2}\], \[3\] moles of \[{H_2}\] and \[0\] moles of ammonia are present. Let the number of moles of gases not present in equilibrium be \[x\]. Then,
Number of moles of \[{N_2} = 1 - x\]
Number of moles of \[{H_2} = 3 - 3x\]
Number of moles of \[N{H_3} = 2x\]
Also, in the equilibrium state, \[0.6\] moles of \[{N_2}\] is present. Hence,
Number of moles of \[{N_2} = 1 - x = 0.6\]
\[x = 0.4\]
Number of moles of \[{H_2} = 3 - 3 \times 0.4 = 3 - 1.2 = 1.8\]
Number of moles of \[N{H_3} = 2x = 2 \times 0.4 = 0.8\]
Therefore, \[0.4\] moles of nitrogen and \[1.8\] moles of hydrogen react to form \[0.8\] moles of ammonia.
Total number of moles of gases present in the flask is \[ = 1 - x + 3 - 3x + 2x\]
\[ = 4 - 2x\]
On putting the value of \[x\], we get,
\[ = 4 - 2 \times 0.4\]
\[ = 4 - 0.8\]
\[ = 3.2\]
Hence, the total number of moles of all the gases present in the flask is \[3.2\].
Hence, Option D is correct.
Note:
When solving the individual equation of gases, students might replace the arbitrary value by \[0.6\]. This will give the correct result for the first equation but it will give an incorrect result for the rest of the equation. And finally adding all the equations will result in an incorrect result. The final addition is also necessary as we are asked for the moles of all the gases present in the flask.
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