Question

1. consider the following reaction:

mathrm{hconh}_{2}(mathrm{g})=mathrm{nh}_{3}(mathrm{g})+mathrm{co}(mathrm{g}) quad k_{c}=4.84 \text { at } 400 mathrm{~k}
if ( 0.186 mathrm{~mol} mathrm{hconh}_{2}(mathrm{g}) ) dissociates in a ( 2.16 mathrm{~l} ) flask at ( 400 mathrm{~k} ), what will be the total pressure in the flask at equilibrium?

Explanation:

Step1: Calculate initial concentration of $HCONH_2$

$C_{HCONH_2}=\frac{n}{V}=\frac{0.186\ mol}{2.16\ L}\approx0.0861\ mol/L$

Step2: Set up the ICE - table

Let $x$ be the change in concentration of $HCONH_2$.

	$HCONH_2$	$NH_3$	$CO$
Change (M)	$-x$	$x$	$x$
Equilibrium (M)	$0.0861 - x$	$x$	$x$

Step3: Write the equilibrium - constant expression

$K_c=\frac{[NH_3][CO]}{[HCONH_2]}$
$4.84=\frac{x\cdot x}{0.0861 - x}$
$4.84(0.0861 - x)=x^{2}$
$0.4167 - 4.84x=x^{2}$
$x^{2}+4.84x - 0.4167 = 0$

Step4: Solve the quadratic equation

Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 1$, $b = 4.84$, and $c=-0.4167$.
$x=\frac{-4.84\pm\sqrt{(4.84)^{2}-4\times1\times(-0.4167)}}{2\times1}$
$x=\frac{-4.84\pm\sqrt{23.4256 + 1.6668}}{2}$
$x=\frac{-4.84\pm\sqrt{25.0924}}{2}$
$x=\frac{-4.84\pm5.0092}{2}$
We take the positive root $x=\frac{-4.84 + 5.0092}{2}=\frac{0.1692}{2}=0.0846\ mol/L$

Step5: Calculate the total concentration at equilibrium

$C_{total}=(0.0861 - x)+x + x=0.0861+x$
Substitute $x = 0.0846\ mol/L$ into the above equation, $C_{total}=0.0861+0.0846=0.1707\ mol/L$

Step6: Use the ideal gas law to find the pressure

The ideal gas law is $PV = nRT$, or $P = CRT$ (where $C=\frac{n}{V}$).
$R = 0.0821\ L\cdot atm/(mol\cdot K)$ and $T = 400\ K$
$P=(0.1707\ mol/L)\times(0.0821\ L\cdot atm/(mol\cdot K))\times400\ K$
$P = 5.62\ atm$

Answer:

$5.62\ atm$