A #3 L# container holds #16 # mol and #24 # mol of gasses A and B, respectively. Groups of five of molecules of gas B bind to three molecules of gas A and the reaction changes the temperature from #340^oK# to #480^oK#. How much does the pressure change?

Question

A #3 L# container holds #16 # mol and #24 # mol of gasses A and B, respectively.  Groups of five of molecules of gas B bind to three molecules of gas A and the reaction changes the temperature from #340^oK# to #480^oK#. How much does the pressure change?

Adrian Ayala · Accepted Answer

The pressure after the reaction will be approximately 0.226 times the pressure before the reaction.

The general plan will be to compute the amount of gas before and after the reaction in moles, and then utilize the ideal gas law to compute the pressure change. We will presume 1) The container's volume remains constant because it is closed and rigid. 2) The response proceeds to its completion and is irreversible. Our response is 3A + 5B #rarr# C Please #n_A^o=# the number of moles of A before the reaction = 16. #n_B^o=# the number of moles of B before the reaction = 24. #n_C^o=# the number of moles of C before the reaction = 0. #n_o=# the total number of moles of gas before the reaction. Take note of that #n_o=n_A^(o)+n_B^(o)+n_C^o=16+24+0=40#. Additionally, let #n_A=# the number of moles of A after the reaction. #n_B=# the number of moles of B after the reaction. #n_C=# the number of moles of C before the reaction. #n=# the total number of moles of gas after the reaction. Take note of that #n=n_A+n_B+n_C#. Assuming that B is the limiting reagent, 24 moles of B would be needed in order to first identify the limiting reagent. 14.4 moles of A are equal to 24 moles B * (5 moles B/3 moles A). B is the limiting reagent because ALL of the B will react with A because we have 16 moles of A instead of 14.4. Following the response, we will have #n_A=n_A^(o)-14.4=16-14.4=1.6# moles of A left, #n_B=0# moles of B left, and #n_C=# 24 moles B * (1 mole C)/(5 moles B) = 4.8 moles C. So #n=n_A+n_B+n_C=1.6+0+4.8=6.4# total moles. Now we use the ideal gas law to determine the pressure change. If #P# = the pressure after the reaction, and #P_o# = the pressure before the reaction, then #P/P_o=(n/n_o)(T/T_o)=(6.4/40)(480/340)~~0.226#

Parker Abbott · Answer

undefined To solve this problem, we can use the ideal gas law, which states that $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is temperature.

First, we calculate the initial pressure using the ideal gas law:

$P_1 = \frac{{n_{\text{total}}RT_1}}{{V}}$

Then, we calculate the final pressure using the same formula but with the new temperature:

$P_2 = \frac{{n_{\text{total}}RT_2}}{{V}}$

Finally, we find the change in pressure:

$\Delta P = P_2 - P_1$

Substituting the given values:

$P_1 = \frac{{(16 + 24)\text{ mol} \times 0.0821 \text{ atm} \cdot \text{L/mol} \cdot 340 \text{ K}}}{{3 \text{ L}}}$

$P_2 = \frac{{(16 + 24)\text{ mol} \times 0.0821 \text{ atm} \cdot \text{L/mol} \cdot 480 \text{ K}}}{{3 \text{ L}}}$

After calculating $P_1$ and $P_2$, we subtract $P_1$ from $P_2$ to find $\Delta P$, the change in pressure.