Question

At high temperatures one mole of hydrogen gas reacts with one mole of bromine gas to form hydrogen bromide. At a given temperature the equilibrium constant is 57.6. If at the same temperature, a mixture of 4.67 × 10^-3M bromine gas, 2.14 × 10^−3 hydrogen gas, and 2.40 × 10^−2M hydrogen bromide gas is made, then:

a. the system is at equilibrium.
b. the system is far from equilibrium and will shift to form more hydrogen gas.
c. the system is far from equilibrium and will shift to form more hydrogen bromide gas.
d. nothing can be deduced since we do not know whether the reaction is endothermic or exothermic.
e. nothing can be deduced since we do not know whether the equilibrium constant is Kc or Kp.

Answer 1

a. the system is at equilibrium.

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`H_2+Br_2\rightleftharpoons 2HBr`

Thus, the law of mass action is given by:

`Keq=([HBr]^2)/([H_2][Br_2]) =57.6`

Nonetheless, for the given point of 4.67 × 10^-3M bromine gas, 2.14 × 10^−3 hydrogen gas, and 2.40 × 10^−2M hydrogen bromide gas we should compute the reaction quotient in order to know whether the direction of the reaction is to left or to right, thus:

`Q=([HBr]^2)/([H_2][Br_2]) =((2.40x10^(-2))^2)/((4.67x10^(-3))(2.14x10^(-3))) \\\\Q=57.6`

Therefore, since Keq=Q, we say that the system is at equilibrium, for that reason, the answer is a.

Best regards.