Question

A mixture in which the mole ratio of hydrogen to oxygen is (exactly) 2:1 is used to prepare water by the reaction 2 H2 (g) + O2 (g) → 2 H2O (g) The total pressure in the container is 0.951 atm at 22°C before the reaction. What is the final pressure in the container after the reaction, with a final temperature of 125°C, no volume change, and an 85.0% yield?

Answer 1

`P_2=0.926atm`

Step-by-step explanation:

Hello,

In this case, it is convenient to state the ideal gas law in terms of the undergoing change due to the chemical reaction:

`(P_1V_1)/(n_1T_1) =(P_2V_2)/(n_2T_2)`

Since no volume change is present, one obtains:

`(P_1)/(n_1T_1) =(P_2)/(n_2T_2)`

Thus, it is possible to assume that the moles of hydrogen and oxygen are 2 mole and 1 mole respectively, in order to respect the given ratio, thus, the yielded moles of water turns out:

`n_(H_2O)=1mol O_2*(2molH_2O)/(1molO_2) =2molH_2O`

It is important to notice that just the 85.0% of the theoretical moles are actually produced, thus:

`n_(H_2O)^(real)=2molH_2O*0.85=1.70molH_2O`

In such a way, there is a remaining amount of hydrogen and oxygen that are computed via:

`n_(H_2)^(remaining)=(2-1.7)molH_2O*(2molH_2)/(2molH_2O)=0.30molH_2\\n_(O_2)^(remaining)=(2-1.7)molH_2O*(1molO_2)/(2molH_2O)=0.15molH_2\\`

Thus, since the initial volume is:

`V=(n_TRT)/(P_1)=((2+1)mol*0.082(atm*L)/(mol*K)*(22+273.15)K)/(0.951atm)= 76.31L`

And it is conserved, at the new temperature, the pressure is:

`P_2=((1.7+0.3+0.15)mol*0.082(atm*L)/(mol*K)(125+273.15)K)/(76.31L) \\\\P_2=0.926atm`

Best regards.