# Question 67d90

Apr 10, 2017

7540 torr

#### Explanation:

K_p =(P_(NOBr) )^2/[(P_(Br_2))(P_(NO))^2 #
The exponents were determined from the coefficients in front of each product or reactant. Also the equilibrium constant is $\frac{P r o \mathrm{du} c t s}{R e a c \tan t s}$ as can be seen with how the equilibrium constant is written. Now we just plug in and solve for the partial pressure of NOBr

$28.4 = {\left({P}_{N O B r}\right)}^{2} / \left[\left(154\right) {\left(114\right)}^{2}\right]$

${P}_{N O B r}$=7539.185 torr

Rounded according to significant figures: 7540 torr