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3 years ago

What is the significance of "Er" in the diagram?

Chemistry

2 answers:

vladimir2022 [97]3 years ago

8 0

Answer:

The significance of "Er" in the diagram is :

B.) Threshold energy for reaction

Explanation:

Threshold energy : It is total amount of energy required by the reactant molecule to reach the transition state .

Activation energy : It is the excess energy absorbed by the molecules to reach the transition state.

Activation Energy = Threshold Energy - Average Kinetic Energy

This means Activation energy decreases on increasing kinetic energy

On increasing Temperature average kinetic energy of the molecule increases which reduces the activation energy and the reaction occur faster in that case.

Catalyst also reduces the Activation energy.

Explanation:

Drupady [299]3 years ago

3 0

Answer:

The significance of "Er" in the diagram is :

B.) Threshold energy for reaction

Explanation:

Threshold energy : It is total amount of energy required by the reactant molecule to reach the transition state .

Activation energy : It is the excess energy absorbed by the molecules to reach the transition state.

Activation Energy = Threshold Energy - Average Kinetic Energy

This means Activation energy decreases on increasing kinetic energy

On increasing Temperature average kinetic energy of the molecule increases which reduces the activation energy and the reaction occur faster in that case.

Catalyst also reduces the Activation energy.

Er = Threshshold energy for reaction at 30 degree

Ea = Activation Energy

The given figure shows that the threshold energy decreases on increasing the temperature

Only the molecule having energy greater than Er can react to form product

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Consider the following reaction: NO( g) + SO 3( g) ⇌ NO 2( g) + SO 2( g) A reaction mixture initially contains 0.86 atm NO and 0

DIA [1.3K]

Answer:

The equilibrium pressure of NO2 is 0.084 atm

Explanation:

Step 1: Data given

A reaction mixture initially contains 0.86 atm NO and 0.86 atm SO3.

Kp = 0.0118

Step 2: The balanced equation

NO( g) + SO3( g) ⇌ NO2( g) + SO2( g)

Step 3: The initial pressures

p(NO) = 0.86 atm

p(SO3) = 0.86 atm

p(NO2) = 0 atm

p(SO2) = 0 atm

Step 4: The pressure at the equilibrium

For 1 mol NO we need 1 mol SO3 to produce 1 mol NO2 and 1 mol SO2

p(NO) = 0.86 -x atm

p(SO3) = 0.86 -xatm

p(NO2) = x atm

p(SO2) = x atm

Step 5: Define Kp

Kp = ((pNO2)*(pSO2)) / ((pNO)*(pSO3))

Kp = 0.0118 = x²/(0.86 - x)²

X = 0.08427

p(NO) = 0.86 -0.08427 = 0.77573 atm

p(SO3) = 0.86 -0.08427 = 0.77573 atm

p(NO2) = 0.08427 atm

p(SO2) = 0.08427 atm

The equilibrium pressure of NO2 is 0.08427 atm ≈ 0.084 atm

5 0

3 years ago

Pls help me I don’t know what to dooooo

Ivan

Molarity = moles of solute/volume of solution in liters.

The solute here is NaCl, of which we have 46.5 g. To calculate the molarity of an NaCl solution, we need to know the number of moles of NaCl. To convert from grams to moles, we divide the mass by the molar mass of NaCl. The molar mass of NaCl is the sum of the atomic masses of Na and Cl: 23 amu + 35 amu = 58 amu. For our purposes, we can regard amu as equivalent to grams/mole.

(46.5 g)/(58 g/mol) = 0.8017 moles NaCl.

Now that we know both the number of moles of our NaCl solute and the volume of the solution, we can calculate the molarity:

(0.8017 moles NaCl)/(2.2 L) = 0.364 M.

5 0

3 years ago

Which states of matter only appear in the hydrosphere

Oliga [24]

I would say the answer is liquids

7 0

3 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago

Determine how many formula units are in 2.35 moles of lithium phosphite?

Stells [14]

Answer:

Determine how many formula units are in 2.35 moles of lithium phosphite?

Explanation:

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4 0

3 years ago