Answer
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:
$$ P_{NOBr} = \sqrt[2]{(28.4) \times {( 0.142 )^{ 2 }( 0.166 )}{}}$$
$$P_{NOBr} = 0.308 \space atm \times \frac{760 \space torr}{1 \space atm} = 234 \space torr$$
