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

Answer:

$P_2=0.926atm$

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:

$\frac{P_1V_1}{n_1T_1} =\frac{P_2V_2}{n_2T_2}$

Since no volume change is present, one obtains:

$\frac{P_1}{n_1T_1} =\frac{P_2}{n_2T_2}$

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:

$n_{H_2O}=1mol O_2*\frac{2molH_2O}{1molO_2} =2molH_2O$

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:

$n_{H_2}^{remaining}=(2-1.7)molH_2O*\frac{2molH_2}{2molH_2O}=0.30molH_2\\n_{O_2}^{remaining}=(2-1.7)molH_2O*\frac{1molO_2}{2molH_2O}=0.15molH_2$

Thus, since the initial volume is:

$V=\frac{n_TRT}{P_1}=\frac{(2+1)mol*0.082\frac{atm*L}{mol*K}*(22+273.15)K}{0.951atm}= 76.31L$

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

$P_2=\frac{(1.7+0.3+0.15)mol*0.082\frac{atm*L}{mol*K}(125+273.15)K}{76.31L} \\\\P_2=0.926atm$

Best regards.