Question

NH4+ (aq) + NO2- (aq) → N2 (g) + H2O (l) Experiment [NH4+]i [NO2-]i Initial rate (M/s) 1 0.24 0.10 7.2 x 10-4 2 0.12 0.10 3.6 x 10-4 3 0.12 0.15 5.4 x 10-4 4 0.12 0.12 4.3 x 10-4 First determine the rate law and rate constant. Under the same initial conditions as in Experiment 4, calculate [NH4+] at 274 seconds after the start of the reaction. In this experiment, both reactants are present at the same initial concentration. The units should be M, and should be calculated to three significant figures.

Answer 1

Answer :

The rate law becomes:

$\text{Rate}=k[NH_4^+][NO_2^-]$

The value of the rate constant 'k' for this reaction is $3.0\times 10^{-2}M^{-1}s^{-1}$

The concentration of $[NH_4^+]$ at 274 seconds after the start of the reaction is 0.0604 M

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:

$NH_4^+(aq)+NO_2^-(aq)\rightarrow N_2(g)+H_2O(l)$

Rate law expression for the reaction:

$\text{Rate}=k[NH_4^+]^a[NO_2^-]^b$

where,

a = order with respect to $NH_4^+$

b = order with respect to $NO_2^-$

Expression for rate law for first observation:

$7.2\times 10^{-4}=k(0.24)^a(0.10)^b$ ....(1)

Expression for rate law for second observation:

$3.6\times 10^{-4}=k(0.12)^a(0.10)^b$ ....(2)

Expression for rate law for third observation:

$5.4\times 10^{-4}=k(0.12)^a(0.15)^b$ ....(3)

Expression for rate law for fourth observation:

$4.3\times 10^{-4}=k(0.12)^a(0.12)^b$ ....(4)

Dividing 2 from 1, we get:

$\frac{7.2\times 10^{-4}}{3.6\times 10^{-4}}=\frac{k(0.24)^a(0.10)^b}{k(0.12)^a(0.10)^b}\\\\2=2^a\\a=1$

Dividing 2 from 3, we get:

$\frac{5.4\times 10^{-4}}{3.6\times 10^{-4}}=\frac{k(0.12)^a(0.15)^b}{k(0.12)^a(0.10)^b}\\\\1.5=1.5^b\\b=1$

Thus, the rate law becomes:

$\text{Rate}=k[NH_4^+]^1[NO_2^-]^1$

$\text{Rate}=k[NH_4^+][NO_2^-]$

Now, calculating the value of 'k' by using any expression.

$7.2\times 10^{-4}=k(0.24)(0.10)$

$k=3.0\times 10^{-2}M^{-1}s^{-1}$

Hence, the value of the rate constant 'k' for this reaction is $3.0\times 10^{-2}M^{-1}s^{-1}$

Now we have to calculate the $[NH_4^+]$ at 274 seconds after the start of the reaction.

For experiment 4 , the initial concentration of $[NH_4^+]$ and $[NO_2^-]$ are same  and the reaction is 1st order for both.

So, it can be considered as a 2nd order reaction.

The expression for second order reaction is:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

where,

k = rate constant = $3.0\times 10^{-2}M^{-1}s^{-1}$

t = time = 274 s

$[A_t]$ = final concentration = ?

$[A_o]$ = initial concentration = 0.12 M

Now put all the given values in the above expression, we get:

$3.0\times 10^{-2}\times 274=\frac{1}{A_t}-\frac{1}{0.12}$

$[A_t]=0.0604M$

Therefore, the concentration of $[NH_4^+]$ at 274 seconds after the start of the reaction is 0.0604 M

Answer 2

The rate constant of this chemical reaction under the given initial conditions is equal to 0.03 M/s.

How to determine the rate law and rate constant.

First of all, we would write the properly balanced chemical equation for this chemical reaction:

                           $NH_4^{+} (aq) + NO_2^{-} (aq) \rightarrow N_2 (g) + H_2O (l)$

Mathematically, the rate law is given by this formula:

$R = K[A]^x[B]^y[C]^z$

Where:

• k is the rate constant.

• A is the concentration of reactant A.

• B is the concentration of reactant B.

• C is the concentration of reactant C.

• x, y, and z are the order of the reaction.

In this scenario, the rate law for this chemical reaction is given by:

$R = K[NH_4^+]^x[NO_2^-]^y$

For the first experiment, we have:

$7.2 \times 10^{-4}=k[0.24]^x[0.10]^y$

For the second experiment, we have:

$3.6 \times 10^{-4}=k[0.12]^x[0.10]^y$

For the third experiment, we have:

$5.4 \times 10^{-4}=k[0.12]^x[0.15]^y$

For the fourth experiment, we have:

$4.3 \times 10^{-4}=k[0.12]^x[0.12]^y$

Next, we would determine the value of x:

$\frac{7.2 \times 10^{-4}}{3.6 \times 10^{-4}} =\frac{k[0.24]^x[0.10]^y}{k[0.12]^x[0.10]^y} \\\\2=2^x$

x = 1.

Also, we would determine the value of y:

$\frac{5.4 \times 10^{-4}}{3.6 \times 10^{-4}} =\frac{k[0.12]^x[0.15]^y}{k[0.12]^x[0.10]^y} \\\\1.5=1.5^x$

y = 1.

Next, we would write the rate law and solve for k as follows:

$R = K[NH_4^+]^1[NO_2^-]^1\\\\7.2 \times 10^{-4}=k[0.24]^1[0.10]^1\\\\7.2 \times 10^{-4}=k[0.24 \times 0.10]\\\\k=\frac{7.2 \times 10^{-4}}{0.024}$

k = 0.03 M/s.

From experiment 4, we can deduce that the chemical reaction is a second order reaction and it is given by this mathematical expression:

$kt=[\frac{1}{A_t} -\frac{1}{A_0}]\\\\0.03 \times 274 =[\frac{1}{A_t} -\frac{1}{0.12}]\\\\8.22=\frac{1}{A_t} -\frac{1}{0.12}\\\\\frac{1}{A_t}=8.22+8.33\\\\\frac{1}{A_t}=16.55\\\\A_t=\frac{1}{16.55} \\\\A_t=0.06\;M$

After the start of the reaction, the concentration of [$NH_4^+$] at 274 seconds is equal to 0.06 M.

Read more on rate constant here: brainly.com/question/24749252