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

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

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

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 :

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

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally