Answer:
of the stock solution would be required.
Explanation:
Assume that a solution of volume contains a solute with a concentration of . The quantity of that solute in this solution would be:
.
For the solution that needs to be prepared, . The volume of this solution is . Calculate the quantity of the solute (magnesium chloride) in the required solution:
.
Rearrange the equation to find an expression of volume , given the concentration and quantity of the solute:
.
Concentration of the solute in the stock solution: .
Quantity of the solute required: .
Calculate the volume of the stock solution that would contain the required of the magnesium chloride solute:
.
Explanation:
H2 (9) + 2 NOg) N20() + H20G)
H2 (M) NO (M) Rate (M*s)
Trial 1 0.30 0.35 2.835 x 10-3
Trial 2 0.60 0.35 1.134 x 10-2
Trial 3 0.60 0.70 2.268 x 10-2
a. What is the order with respect to H2?
Comparing trials 1 and 2,the concentration of H is doubled and that leads to an increase in the rate of the reaction by a factor of 4. This means the order with respect to H is 2.
b. What is the order with respect to NO?
Comparing trials 2 and 3, the concentration of NO is doubled and that leads to an increase in the rate of the reaction by a factor of 2. This means the order with respect to NO is 1.
c. What is the rate equation for this reaction?
rate = k [H]²[NO]
d. Calculate the rate constant for the reaction.
Taking trial 1;
rate = k [H]²[NO]
2.835 x 10-3 = k (0.30)² (0.35)
k = 90 x 10-3 = 0.09 L2 mol-2 s-1