## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 15 - Exercises - Page 715: 40

#### Answer

$$0.0176 \space torr$$

#### Work Step by Step

- Convert the pressures to atm: $$137 \space torr \times \frac{1 \space atm}{760 \space torr} = 0.180 \space atm$$ $$285 \space torr \times \frac{1 \space atm}{760 \space torr} = 0.375 \space atm$$ 1. The exponent of each concentration is equal to its balance coefficient. $$K_P = \frac{[Products]}{[Reactants]} = \frac{P_{ SO_2 }P_{ Cl_2 }}{P_{ SO_2Cl_2 }}$$ 2. Solve for the missing concentration: $$P_{SO_2Cl_2} = \frac{P_{ SO_2 } P_{ Cl_2 } }{ K_P}$$ 3. Evaluate the expression: $$P_{SO_2Cl_2} = \frac{( 0.18 )( 0.375 )}{(2.91 \times 10^3)}$$ $$[SO_2Cl_2] = 2.32 \times 10^{-5} \space atm \times \frac{760 \space torr}{1 \space atm} = 0.0176 \space torr$$

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