# Chapter 15 - Exercises - Page 715: 39

The partial pressure of NOBr in this mixture is equal to 234 torr.

#### Work Step by Step

- Convert the pressures to atm: $$108 \space torr \times \frac{1 \space atm}{760 \space torr} = 0.142 \space atm$$ $$126 \space torr \times \frac{1 \space atm}{760 \space torr} = 0.166 \space atm$$ - The exponent of each concentration is equal to its balance coefficient. $$K_P = \frac{[Products]}{[Reactants]} = \frac{P_{ NOBr } ^{ 2 }}{P_{ NO } ^{ 2 }P_{ Br_2 }}$$ 2. Solve for the missing concentration: $$\sqrt[2]{K_P \times P_{ NO } ^{ 2 }P_{ Br_2 }}{} = P_{NOBr}$$ 3. Evaluate the expression: $$P_{NOBr} = \sqrt[2]{(28.4) \times {( 0.142 )^{ 2 }( 0.166 )}{}}$$ $$P_{NOBr} = 0.308 \space atm \times \frac{760 \space torr}{1 \space atm} = 234 \space torr$$

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