The rate of a reaction would be one-fourth.
<h3>Further explanation</h3>
Given
Rate law-r₁ = k [NO]²[H2]
Required
The rate of a reaction
Solution
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.
Can be formulated:
Reaction: aA ---> bB

or

The concentration of NO were halved, so the rate :
![\tt r_2=k[\dfrac{1}{2}No]^2[H_2]\\\\r_2=\dfrac{1}{4}k.[No]^2[H_2]\\\\r_2=\dfrac{1}{4}r_1](https://tex.z-dn.net/?f=%5Ctt%20r_2%3Dk%5B%5Cdfrac%7B1%7D%7B2%7DNo%5D%5E2%5BH_2%5D%5C%5C%5C%5Cr_2%3D%5Cdfrac%7B1%7D%7B4%7Dk.%5BNo%5D%5E2%5BH_2%5D%5C%5C%5C%5Cr_2%3D%5Cdfrac%7B1%7D%7B4%7Dr_1)
25/2 and 96/X
CROSS MULTIPLY.
2x=2,400.
divide by 2.
x=1,200.
you take the GIVEN MASS of an element, and you put it on top, the coefficient is what it’s over. i believe this is right
Answer:
CH3OH + 02 ----> C02 + H20
balanced equation -
CH3OH + 3/202 ----> C02 + 2H20
Use exactly the same process as the one used on another question of yours I answered.