The equilibrium constants for the reaction Br2 leftharpoons 2Br at 500 K and 700 K are 1 × 10-10 and 1 × 10-5 respectively. The reaction is:

Question

The equilibrium constants for the reaction Br2 leftharpoons 2Br at 500 K and 700 K are 1 × 10-10 and 1 × 10-5 respectively. The reaction is: (A) Endothermic (B) Exothermic (C) Fast (D) Slow

Tardigrade Admin · Accepted Answer

Chemical equilibrium constant for the reaction is

B r_{2} ⇌ 2 B r

is

K_{c} = \frac{[ B r ] ^{2}}{[ B r _{2} ]}

The value of

K_{c}

at

500 K

is

1 \times 1 0^{- 10}

. One increasing temperature

(700 K) .

The value of

K_{c}

is also increased, i.e., the concentration of the product is increased and obviously the increase in temperature will favour the forward reaction.

Q. The equilibrium constants for the reaction Br2​⇌2Br at 500K and 700K are 1×10−10 and 1×10−5 respectively. The reaction is:

Endothermic

Exothermic

Fast

Slow

Q. The equilibrium constants for the reaction $B r_{2} ⇌ 2 B r$ at $500 K$ and $700 K$ are $1 \times 1 0^{- 10}$ and $1 \times 1 0^{- 5}$ respectively. The reaction is: