Question

Butane gas burns in air according to the following reaction,

$2C_4{H}_{10}+13O_2⟶10H_{2}O+8{CO}_2$.

Suppose the initial and final temperatures are equal and high enough so that all reactants and products act as perfect gases. Two moles of butane are mixed with $13$ moles of oxygen and then completely reached. Find the final pressure (if the volume remains unchanged and the pressure before the reaction is $P_0$)?

Answer 1

$2C_4{H}_{10}+13O_2\to 10H_{2}O+8C{O}_2$

Pressure before reaction $=P_0$

$2C_4{H}_{10}+13O_2\to 10H_{2}O+8C{O}_2$

At $t=0$ 2 moles 13 moles

At $t=t(2-2)(13-13)10moles8moles$

= 0 mol = 0 mol of $H_{2}O\left(g\right)$ of $C{O}_2$

At constant volume and temperature :$\to$

$PV=nRT$

$\frac{P}{n}=\frac{RT}{V}$ constant

$\frac{P}{n}=$ constant

$\frac{P_1}{n_1}=\frac{P_2}{n_2}$

So, As initial pressure was $P_0$

$P_1=P_0$

initial moles $n_1=15$ moles $(2+13)$

final mole $n_2=18$ mole $(10+8)$

Now,

$\frac{P_0}{15}=\frac{P_2}{18}\Rightarrow P_2=\frac{6}{5}{P}_0$