A #7 L# container holds #28 # mol and #12 # mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to five molecules of gas A and the reaction changes the temperature from #150^oK# to #160^oK#. How much does the pressure change?

Question

A #7 L# container holds #28 # mol and #12 # mol of gasses A and B, respectively.  Groups of three of molecules of gas B bind to five molecules of gas A and the reaction changes the temperature from #150^oK# to #160^oK#. How much does the pressure change?

Emily Abbate · Accepted Answer

All of the gas A and some of the gas B will react to produce #A_3B_5#. The final pressure will change due to the change in the temperature and the change in the number of moles of gas present.

The pressure will change by #2288# #kPa#.

(As we are not informed otherwise, we must presume that the gases are monatomic prior to the reaction.) The response will be: #3A + 5B -> A_3B_5# In other words, 1 mol of product gas is produced for every 8 mol of reactant gases combined. Since the gases are not present in this ratio, after the reaction one will be "in excess" of the other. There is much more of gas A than gas B, but we need more of gas B for each reaction, so all the gas B will be used up and some gas A will be left over. After the reaction, the container will include product gas #A_3B_5# and gas A. The reaction in the equation above will occur #12/5 = 2.4# times, so there will be #2.4# mol of #A_3B_5#. #2.4xx3=7.2# mol of gas A will react, so #28-7.2 = 20.8# mol of gas A will remain. That means there will be a total of #20.8+2.4 = 23.2# mol of gas in the container in total after the reaction. There were #28+12=40# mol of gas in the container before the reaction. The universal gas equation can be applied here: #PV=nRT# Organizing: #P=(nRT)/V# Since the volume of the container (V) and the gas constant (R) in this instance are both constant, we can eliminate them from the expression: #P_1 \prop n_1T_1 = 40xx150=6000# #P_2 \prop n_2T_2 = 23.2xx160=3712# The pressure will change by #2288# #kPa#.

Elena Albarran · Answer

undefined Use the ideal gas law to find the initial and final pressures:

$$ P_1 = \frac{n_1RT_1}{V} $$

$$ P_2 = \frac{n_2RT_2}{V} $$

where $P$ is pressure, $n$ is the number of moles, $R$ is the ideal gas constant, $T$ is temperature in Kelvin, and $V$ is volume.

Calculate initial pressure ($P_1$) using the initial conditions, and then calculate final pressure ($P_2$) using the final conditions.

The pressure change ($\Delta P$) is given by:

$$ \Delta P = P_2 - P_1 $$