For a reaction that follows the general rate law, Rate = k[A][B]2, what will happen to the rate of reaction if the concentration of B is increased by a factor of 4.00? The rate will

Answer : The rate is increased by 16 times.

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The given rate law expression is:

As per question, when the concentration of B increased by factor of 4.00 then the rate law expression will be:

From this we conclude that, when the concentration of B increased by factor of 4.00 then the rate is increased by 16 times.

Thus, the rate is increased by 16 times.

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

b is the correct