Question

The compound SbCl₅(g) decomposes at high temperatures to gaseous antimony trichloride and chlorine gas. When 89.7 g of SbCl₅(g) is placed in a 15.0-L container at 180⁰C, the SbCl₅(g) is 29.2% decomposed (by moles) after the system has reached equilibrium.

a. Calculate the value of K for this reaction at 180⁰C.

b. Determine the number of moles of chlorine gas that must be injected into the flask to make the new equilibrium pressure of antimony trichloride half that of the original equilibrium pressure of antimony trichloride in the original experiment.

Short answer

(a) The value of K for this reaction at 180⁰C will be $2.41\times {10}^{-3}mol/L$

(b) 0.168 mol of Cl₂ was injected.

Step-by-step solution

Determine the equilibrium constant K

(a)

No of moles of SbCl₅ present initially

$89.7gSbC{l}_5\times \frac{1mol}{299.1g}=0.3molSbC{l}_5$

The reaction will be as follows

$$\begin{gathered} 0.3\left(0.292\right)=0.0876 \\ 0.3-0.3\left(0.292\right)=0.212 \end{gathered}$$

Therefore, the equilibrium constant (K) will be

$\begin{array}{rcl}K& =& \frac{\left[SbC{l}_3\right]\left[C{l}_2\right]}{\left[SbC{l}_5\right]}\\ & =& \frac{\left({\displaystyle \frac{0.0876mol}{15L}}\right)\left({\displaystyle \frac{0.0876mol}{15L}}\right)}{\left({\displaystyle \frac{0.212mol}{15L}}\right)}\\ & =& 2.41\times {10}^{-3}mol/L\end{array}$

Determine the number of moles of Cl2 injected

Let x be the no of moles of Cl₂ added

It has been stated that reaction shifts left after addition of Cl₂

Moles of SbCl₃ at equilibrium

$$\begin{gathered} =\frac{0.0876}{2} \\ =0.0438mol \end{gathered}$$

Moles of Cl₂ at equilibrium

role="math" localid="1659962695862" $$\begin{gathered} =x+0.0876-0.0438 \\ =x+0.0438 \end{gathered}$$

Moles of SbCl₅ at equilibrium

$$\begin{gathered} =0.212+0.0438 \\ =0.256 \end{gathered}$$

Now we have to calculate the amount of Cl₂ added.

role="math" localid="1659962907643" $\begin{array}{rcl}K& =& \frac{\left[SbC{l}_3\right]\left[C{l}_2\right]}{\left[SbC{l}_5\right]}\\ 2.41\times {10}^{-3}mol/L& =& \frac{\left({\displaystyle \frac{x+0.0438mol}{15L}}\right)\left({\displaystyle \frac{0.0438mol}{15L}}\right)}{\left({\displaystyle \frac{0.256mol}{15L}}\right)}\\ x& =& 0.168\end{array}$

Therefore, 0.168 mol of Cl₂ was added.