Question

Given that the following reaction begins at equilibrium, which of the following describes the reaction immediately after a small portion of CO₂(g)
is removed from the reaction container?

CO(g) + H₂O(g) <----> CO₂(g) + H₂(g)

		Q < Kc, reaction is spontaneous in reverse direction
		Q < Kc, reaction is spontaneous in forward direction
		Q > Kc, reaction is spontaneous in reverse direction
		Q > Kc, reaction is spontaneous in forward direction
		No effect 2. Use the initial concentration and initial rate data in the table to determine the rate law for the reaction:   A(aq) + B(aq)   →   C(aq) + D(g) Experiment # [A] (M) [B] (M) Initial Rate (M/s) 1 0.400 0.400 9.00×10-3 2 0.400 0.200 4.50×10-3 3 0.200 0.200 1.13×10-3 Rate = k[A] Rate = k[B]2 Rate = k[B] Rate = k[A]2[B] Rate = k[A][B]2

Answer 1

Solution :-

Q1)Since CO2 is present at the product side therefore after removing the CO2 the concentration of the product will decrease which will lead to the decrease in the Q value because Q= [product]/[reactant] therefore to attain the new equilibrium the reaction will shift to the right side resulting in the formation of the more CO2

Therefore the correct answer for the reaction is

Q<Kc , reaction is spontaneous in forward direction.

Q2)Solution

Given data

A(aq) + B(aq)   ----- >   C(aq) + D(g)

Determine the rate law for reaction

To determine the rate law of the reaction we need to determine the order of reaction with respect to each reactant and then we can write the overall rate law for the reaction.

So lets calculate the order of the each reactant using the concentrations and the initial rate values

Lets first calculate the order with respect to the reactant A(aq)

To find the order with respect to A we need to select the such experiments where concentration of the A is different and concentrations of B are same so the concentration term of B is get cancelled out and we get the order with respect to A

So lets use the concentrations and rate values from the experiment 2 and 3

So we can make the set up as follows

[rate 2/rate 3] = ([A]2/[A]3)^m

Now lets put the values in the formul a

[4.50*10^-3/1.13*10^-3] = (0.400/0.200)^m

3.98 = 2^m

Therefore

log 3.98 = m * log 2

m = log 3.98 / log 2

m = 2.0

So the order with respect to reactant a is second order

Now lets calculate the order with respect the reactant using the experiment 1 and 2

[rate 1/rate 2] = ([B]1/[B]2)ⁿ

Now lets put the values in the formul a

[9.00*10^-3/4.50*10^-3] = (0.400/0.200)ⁿ

2 = 2ⁿ

Therefore

log 2 = n * log 2

n = log 2 / log 2

n = 1

therefore the order with respect to the reactant B is first order.

So the rate law for the reaction is as follows

Rate = k [A]²[B]