2NaOH + H2SO4 → Na2SO4 + H2O is this equation balanced

Be sure to answer all parts. Consider the reaction A + B → Products From the following data obtained at a certain temperature, d

worty [1.4K]

Answer : The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

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:

$A+B\rightarrow Products$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first observation:

$3.20\times 10^{-1}=k(1.50)^a(1.50)^b$ ....(1)

Expression for rate law for second observation:

$3.20\times 10^{-1}=k(1.50)^a(2.50)^b$ ....(2)

Expression for rate law for third observation:

$6.40\times 10^{-1}=k(3.00)^a(1.50)^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0$

Dividing 1 from 3, we get:

$\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[A]^1[B]^0$

$\text{Rate}=k[A]$

Thus,

The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

7 0

3 years ago

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step me

slega [8]

Answer:

The two-step mechanism is a slow mechanism and a fast mechanism. When we combine them, the result is

2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)

Explanation:

We know that the decomposition of hydrogen peroxide is catalyzed by iodide ion, which means that the iodide ion will react with the hydrogen peroxide. There is a slow mechanism and a fast one:

H₂O₂(aq) + I₋(aq) ⇒ H₂O(l) + IO₋(aq) this is the slow reaction

IO₋(aq) + H₂O₂(aq)⇒ H₂O(l) + O₂(g) + I₋ (aq) this is the fast reaction

If we cancel the same type of molecules and ions, the final result is:

2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)

3 0

4 years ago

Do nonmetals form ions by gaining electrons?

Ray Of Light [21]

The answer is yes they do form ions upon gaining electrons and becomes negatively charged

5 0

4 years ago

For a research project on time management hearing is to calculate how many hours the average high school student spends in schoo

ivolga24 [154]

Answer: it’s b

Explanation:

3 0

4 years ago

Read 2 more answers

The heat of vaporization of water at the normal boiling point, 373.2 K, is 40.66 kJ/mol. The molar heat capacity of liquid water

Tatiana [17]

Answer:

$\Delta _{vap}H(300.2K)=43,658\frac{J}{mol}=43.66\frac{kJ}{mol}$

Explanation:

Hello!

In this case, according to the Kirchhoff's law for the enthalpy change, it is possible to compute the heat of vaporization at 300.2 K by considering the following thermodynamic route:

$\Delta _{vap}H(300.2K)=Cp_{liq}(T_b-T\°)+\Delta _{vap}H\°+Cp_{vap}(T-T_B)$

Whereas the first term stands for the effect of taking the liquid from 298.15 K to 373.15 K, the second term stands for the standard enthalpy of vaporization and the last term that of the vapor from the boiling point to 300.2 K; thus we plug in to obtain:

$\Delta _{vap}H(300.2K)=75.37\frac{J}{mol*K} (373.2K-298.15K)+40,660\frac{J}{mol} +36.4\frac{J}{mol*K}(300.2K-373.2K)\\\\\Delta _{vap}H(300.2K)=43,658\frac{J}{mol}=43.66\frac{kJ}{mol}$

Best regards!

6 0

3 years ago