asked 91.0k views
5 votes
Select all the correct answers.

In this reaction, how does the rate of forward reaction vary with the concentration of the product?
2H,S() =2H() + S(g)
O It increases with an increase in the concentration of S-(g).
OIt decreases with a decrease in the concentration of H().
OIt increases with a decrease in the concentration of H-(g).
It decreases with an increase in the concentration of S,().
It decreases with increase in the concentration of Ha().

asked
User Msg
by
7.7k points

1 Answer

5 votes

The correct answers :

It increases with a decrease in the concentration of H₂(g).

It decreases with an increase in the concentration of S₂(g).

It decreases with an increase in the concentration of H₂(g).

Further explanation

Forward reaction : rate to form product

In equilibrium :

The product decreases ⇒ system will move from left to right(forward reaction)

The product increases ⇒ system will move from right to left(reverse reaction)

answered
User Kubi
by
7.8k points

No related questions found

Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.