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3 years ago

4NH3 (g)+502(g)->6H20g)+4NOg) What is the enthalpy change for this reaction?

1 answer:

7 0

Enthalpy is energy of bonds broken - energy of bonds formed. Here, the NH3 and O2 are broken and H2O and NO are formed. So the energy to break the NH3 bonds is 3 times the amount of energy it takes to break a N-H single bond (because there are three of them in a NH3 molecule) and then multiplied by 4 because there are four particles.

So the energy of the bonds broken is 12x the energy to break a N-H single bond plus 5x the amount of energy to break an O—O double bond (you don’t multiply this by anything because in each O2 molecule there is only one bond).

The energy of the bonds formed is 6*2 = 12 Times the amount of energy for a O-H single bond plus 4 times the amount of energy required to break a N—O double bond.

Subtract energy of bonds broken - energy of bonds formed and this is the change in enthalpy.

To know what type of bond it is, draw the Lewis structure.

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Magnesium oxide (MgO) forms when the metal burns in air. (a) If 1-25 9 of MgO contains 0.754 g of Mg, what is the mass ratio of

Vsevolod [243]

Answer:

Explanation:

a )

1.25 g MgO contains .754 g of Mg .Rest will be O

so oxide = 1.25 - .754 = 0.496 g

ratio of magnesium to oxide = .754/.496 = 1.52

b) 1.25 g of MgO contains .754 g of Mg

534 g of MgO contains .754 x 534 / 1.25 g = 322.11 g

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2 years ago

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NISA [10]

Answer:

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8 0

2 years ago

Calculate the activity coefficients for the following conditions:

uysha [10]

Answer:

For a: The activity coefficient of copper ions is 0.676

For b: The activity coefficient of potassium ions is 0.851

For c: The activity coefficient of potassium ions is 0.794

Explanation:

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

$-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}$ ........(1)

where,

$\gamma_i$ = activity coefficient of ion

$Z_i$ = charge of the ion

$\mu$ = ionic strength of solution

$\alpha _i$ = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

$\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)$ ......(2)

where,

$C_i$ = concentration of i-th ions

$Z_i$ = charge of i-th ions

For a:

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

$C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M$

Now, calculating the activity coefficient of $Cu^{2+}$ ion in the solution by using equation 1:

$Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{ (known)}\\\mu=0.01M$

Putting values in equation 1, we get:

$-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676$

Hence, the activity coefficient of copper ions is 0.676

For b:

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

$C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.025M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851$

Hence, the activity coefficient of potassium ions is 0.851

For c:

We are given:

0.02 M $K_2SO_4$ solution:

Calculating the ionic strength by using equation 2:

$C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.06M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794$