Answer : The rate for trial 5 will be
Explanation :
Rate law 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.
For the given chemical equation:
Rate law expression for the reaction:
where,
a = order with respect to A
b = order with respect to B
c = order with respect to C
Expression for rate law for first observation:
....(1)
Expression for rate law for second observation:
....(2)
Expression for rate law for third observation:
....(3)
Expression for rate law for fourth observation:
....(4)
Dividing 1 from 2, we get:
Dividing 1 from 3, we get:
Dividing 3 from 4, we get:
Thus, the rate law becomes:
Now, calculating the value of 'k' by using any expression.
Putting values in equation 1, we get:
Thus, the value of the rate constant 'k' for this reaction is
Now we have to calculate the rate for trial 5 that starts with 0.90 M of reagent A, 0.60 M of reagents B and 0.70 M of reagent C.
Therefore, the rate for trial 5 will be