Chapter 12, Problem 3QAP

Interpretation Introduction

Interpretation:

For the given reaction, the given table needs to be completed for partial pressures of A, B and C gaseous species.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place. The ICE table is used to determine the partial pressure or concentration of species involved in equilibrium at given time. Here, I stands for initial partial pressure, C stands for change in the partial pressure and E stands for partial pressure at equilibrium. In the given case, the partial pressure of species can be determined using the change in the partial pressure as per the balanced chemical reaction at given time.

Answer to Problem 3QAP

Time (s)	$P_A\left(\text{atm}\right)$	$P_B\left(\text{atm}\right)$	$P_C\left(\text{atm}\right)$
0	1.850	0.900	0.000
10	1.7934	0.8717	0.0850
20	1.500	0.725	0.525
30	1.25	0.600	0.9
40	1.100	0.525	1.125
50	0.920	0.435	1.395
60	0.920	0.435	1.395

Explanation of Solution

The reaction of the system is as follows:

$2\text{A}\left(g\right)+\text{B}\left(g\right)\to 3\text{C}\left(g\right)$

The given incomplete data is as follows:

Time (s)	$P_A\left(\text{atm}\right)$	$P_B\left(\text{atm}\right)$	$P_C\left(\text{atm}\right)$
0	1.850	0.900	0.000
10			0.0850
20	1.500
30		0.600
40	1.100
50		0.435
60	0.920

The partial pressure of all the species at given time can be calculated as follows:

$\begin{array}{l}2\text{A}\left(g\right)+\text{B}\left(g\right)\to 3\text{C}\left(g\right)\\ t=0\text{ 1}\text{.850 0}\text{.900 0}\text{.000}\\ t=10\text{ 1}\text{.850-}2x\text{ 0}\text{.900-}x\text{ 3}x\end{array}$

Since, the value of partial pressure of C at time 10 s is 0.0850 atm.

Thus,

$3x=0.0850$

Or,

$x=\frac{0.0850}{3}=0.0283$

Now, the partial pressure of gas A and B can be calculated as follows:

$P_A=1.850-2x=1.850-2\left(0.0283\right)=1.7934\text{ atm}$

And,

$P_B=0.900-x=0.900-0.0283=0.8717\text{ atm}$

Therefore, the partial pressure of gas A and B at 10 s is 1.7934 atm and 0.8717 atm respectively.

Now, at time 20 s:

$\begin{array}{l}2\text{A}\left(g\right)+\text{B}\left(g\right)\to 3\text{C}\left(g\right)\\ t=0\text{ 1}\text{.850 0}\text{.900 0}\text{.000}\\ t=20\text{ 1}\text{.850-}2x\text{ 0}\text{.900-}x\text{ 3}x\end{array}$

Since, the value of partial pressure of A at time 20 s is 1.500 atm.

Thus,

$1.850-2x=1.500$

Or,

$x=\frac{1.850-1.500}{2}=0.175$

Now, the partial pressure of gas B and C can be calculated as follows:

$P_B=0.900-x=0.900-0.175=0.725\text{ atm}$

And,

$P_C=3x=3\left(0.175\right)=0.525\text{ atm}$

Therefore, the partial pressure of gas B and C at 20 s is 0.725 atm and 0.525 atm respectively.

Similarly, at time 30 s:

$\begin{array}{l}2\text{A}\left(g\right)+\text{B}\left(g\right)\to 3\text{C}\left(g\right)\\ t=0\text{ 1}\text{.850 0}\text{.900 0}\text{.000}\\ t=30\text{ 1}\text{.850-}2x\text{ 0}\text{.900-}x\text{ 3}x\end{array}$

Since, the value of partial pressure of B at time 30 s is 0.600 atm.

Thus,

$0.900-x=0.600$

Or,

$x=0.900-0.600=0.300$

Now, the partial pressure of gas A and C can be calculated as follows:

$P_A=1.850-2x=1.850-2\left(0.300\right)=1.25\text{ atm}$

And,

$P_C=3x=3\left(0.3\right)=0.9\text{ atm}$

Therefore, the partial pressure of gas A and C at 30 s is 1.25 atm and 0.9 atm respectively.

At $t=40\text{ s}$ :

$\begin{array}{l}2\text{A}\left(g\right)+\text{B}\left(g\right)\to 3\text{C}\left(g\right)\\ t=0\text{ 1}\text{.850 0}\text{.900 0}\text{.000}\\ t=40\text{ 1}\text{.850-}2x\text{ 0}\text{.900-}x\text{ 3}x\end{array}$

Since, the value of partial pressure of A at time 40 s is 1.100 atm.

Thus,

$1.850-2x=1.100$

Or,

$x=\frac{1.850-1.100}{2}=0.375$

Now, the partial pressure of gas B and C can be calculated as follows:

$P_B=0.900-x=0.900-0.375=0.525\text{ atm}$

And,

$P_C=3x=3\left(0.375\right)=1.125\text{ atm}$

Therefore, the partial pressure of gas B and C at 20 s is 0.525 atm and 1.125 atm respectively.

Now, at time 50 s:

$\begin{array}{l}2\text{A}\left(g\right)+\text{B}\left(g\right)\to 3\text{C}\left(g\right)\\ t=0\text{ 1}\text{.850 0}\text{.900 0}\text{.000}\\ t=50\text{ 1}\text{.850-}2x\text{ 0}\text{.900-}x\text{ 3}x\end{array}$

Since, the value of partial pressure of B at time 50 s is 0.435 atm.

Thus,

$0.900-x=0.435$

Or,

$x=0.900-0.435=0.465$

Now, the partial pressure of gas A and C can be calculated as follows:

$P_A=1.850-2x=1.850-2\left(0.465\right)=0.92\text{ atm}$

And,

$P_C=3x=3\left(0.465\right)=1.395\text{ atm}$

Therefore, the partial pressure of gas A and C at 50 s is 0.92 atm and 1.395 atm respectively.

Lastly, At $t=60\text{ s}$ :

$\begin{array}{l}2\text{A}\left(g\right)+\text{B}\left(g\right)\to 3\text{C}\left(g\right)\\ t=0\text{ 1}\text{.850 0}\text{.900 0}\text{.000}\\ t=60\text{ 1}\text{.850-}2x\text{ 0}\text{.900-}x\text{ 3}x\end{array}$

Since, the value of partial pressure of A at time 60 s is 0.920 atm.

Thus,

$1.850-2x=0.920$

Or,

$x=\frac{1.850-0.920}{2}=0.465$

Now, the partial pressure of gas B and C can be calculated as follows:

$P_B=0.900-x=0.900-0.465=0.435\text{ atm}$

And,

$P_C=3x=3\left(0.465\right)=1.395\text{ atm}$

Therefore, the partial pressure of gas B and C at 60 s is 0.435 atm and 1.395 atm respectively.

The complete table for partial pressure of all the gases at given time will be:

Time (s)	$P_A\left(\text{atm}\right)$	$P_B\left(\text{atm}\right)$	$P_C\left(\text{atm}\right)$
0	1.850	0.900	0.000
10	1.7934	0.8717	0.0850
20	1.500	0.725	0.525
30	1.25	0.600	0.9
40	1.100	0.525	1.125
50	0.920	0.435	1.395
60	0.920	0.435	1.395

Conclusion

The complete table for partial pressure of all the gases at given time will be:

Time (s)	$P_A\left(\text{atm}\right)$	$P_B\left(\text{atm}\right)$	$P_C\left(\text{atm}\right)$
0	1.850	0.900	0.000
10	1.7934	0.8717	0.0850
20	1.500	0.725	0.525
30	1.25	0.600	0.9
40	1.100	0.525	1.125
50	0.920	0.435	1.395
60	0.920	0.435	1.395