Questions: Consider the following equilibrium at 333 K:
H2S(g) + I2(s) ⇌ S(s) + 2 HI(g)
Kp = 1.34 × 10^-5
H2S at a partial pressure of 786.2 mmHg and 1.68 g of I2 (MM = 253.80 g mol^-1) are added to an otherwise empty 1.00 L flask at 333 K and allowed to come to equilibrium.
Determine the partial pressure of HI in mmHg at equilibrium.
Transcript text: Consider the following equilibrium at 333 K :
\[
\begin{array}{r}
\mathrm{H} 2 \mathrm{~S}(\mathrm{~g})+\mathrm{I} 2(\mathrm{~s}) \rightleftharpoons \mathrm{S}(\mathrm{~s})+2 \mathrm{HI}(\mathrm{~g}) \\
K p=1.34 \times 10^{\wedge}-5
\end{array}
\]
H 2 S at a partial pressure of 786.2 mmHg and 1.68 g of $\mathrm{I} 2(\mathrm{mM}=253.80 \mathrm{~g} \mathrm{~mol}-1)$ are added to an otherwise empty 1.00 L flask at 333 K and allowed to come to equilibrium.
Determine the partial pressure of HI in mmHg at equilibrium.
INSTRUCTIONS: Give your answer to 3 significant digits in scientific notation (example: $1.23 \times 10^{\wedge}-4$ would be typed in as 1.23E-4) in the MS Form
Formatting may be lost in the question above - see the slide shown in class for correct formatting
Enter your answer
Solution
Solution Steps
Step 1: Write the equilibrium expression
The equilibrium constant Kp for the reaction is given by:
Kp=PH2SPHI2
Step 2: Convert initial conditions to appropriate units
Given:
Partial pressure of H2S is 786.2 mmHg.
Mass of I2 is 1.68 g.
Molar mass of I2 is 253.80 g/mol.
Volume of the flask is 1.00 L.
Calculate the initial moles of I2:
moles of I2=253.80g/mol1.68g=0.00662mol