Question

If the rate of formation of ammonia gas in haber bosch 2.5x10-4 mol/l-1/s-1 then rate of consumption of n2 gas during process equals... mol/l-1/s-1

A)-2.5x10-4
B) -1.25x10-4
C)-3.75x10-4
D)-5x10-4
Can somebody help me out i am stressed

Answer 1

Rate of consumption of N₂ = - 1.25 x 10⁻⁴

Further explanation

The reaction rate (v) shows the change in the concentration per unit time.

`\tt rate=(concentration(M))/(time(s))`

The rate reaction for Haber Bosch :

N₂(g)+3H₂(g)⇒2NH₃(g)

From equation, mol ratio N₂ : NH₃ = 1 : 2, so

the rate of formation of ammonia = 2 x the rate of consumption of nitrogen.

Given

the rate of formation of ammonia gas : 2.5x10⁻⁴ M/s

the rate of consumption of nitrogen :

`\tt (2.5* 10^(-4))/(2)=1.25* 10^(-4)`

The sign negative for consumption : -1.25 x 10⁻⁴ M/s

or we can use another way:

Rate of formation of NH₃ = reaction rate x coefficient of NH₃

2.5 x 10⁻⁴ = reaction rate x 2

reaction rate = 1.25 x 10⁻⁴

Rate of consumption of N₂ = reaction rate x coefficient of N₂

Rate of consumption of N₂ = 1.25 x 10⁻⁴ x 1

Rate of consumption of N₂ = 1.25 x 10⁻⁴