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3 years ago

Will medal!!!

1 answer:

hjlf3 years ago

5 0

Δ H reaction = q / n where q: amount of heat released and n is number of moles of substance.
q = m . C . ΔT where:
m = mass of substance (g)
C = Specific heat capacity (4.18)
ΔT = change in temperature = 24.25 - 23.16 = 1.09
q = 1000 x 4.18 x 1.09 = 4556 J = 4.556 kJ
number of moles (n) = Molarity (M) x Volume (L)
= 0.185 M x 0.07 L = 0.01295 mole
Δ H = q / n = - (4.556 kJ / 0.01295 mole) = -351.8 kJ / mol
Note: it is exothermic reaction (-ve sign) i.e. temperature is raised

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3 years ago

Read 2 more answers

The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.

Brums [2.3K]

Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

$\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]$

Where,

$\Delta H_v$ is the Heat of vaoprization (J/mol)

$T_b$ is the normal boiling point of the gas (K)

$T_c$ is the Critical temperature of the gas (K)

$P_c$ is the Critical pressure of the gas (bar)

R is the gas constant (8.314 J/Kmol)

For diethyl ether:

$T_b=307.4\ K$

$T_c=466.7\ K$

$P_c=36.4\ bar$

Applying the above equation to find heat of vaporization as:

$\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]$

$\Delta H_v=26400 J/mol$

The conversion of J into kJ is shown below:

1 J = 10⁻³ kJ

Thus,

$\Delta H_v=26.4 kJ/mol$

Option C is correct

6 0

4 years ago

The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r

spin [16.1K]

Answer: The theoretical yield of acetanilide is 6.5 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For aniline:

Given mass of aniline = $4.50\times 10^0=4.50g$ (We know that: $10^0=1$ )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

$\text{Moles of aniline}=\frac{4.50g}{93.13g/mol}=0.048mol$

For acetic anhydride:

To calculate the mass of acetic anhydride, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Volume of acetic anhydride = $(1.25\times \text{Mass of aniline})=1.25\times 4.50=5.625mL$

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

$1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL\times 5.625mL)=6.08g$

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of acetic anhydride}=\frac{6.08g}{102.1g/mol}=0.06mol$

The chemical equation for the reaction of aniline and acetic anhydride follows:

$C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH$

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with = $\frac{1}{1}\times 0.048=0.048mol$ of acetic anhydride

As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.

Thus, aniline is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce = $\frac{1}{1}\times 0.048=0.048mol$ of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

$0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol\times 135.17g/mol)=6.5g$

Hence, the theoretical yield of acetanilide is 6.5 grams.

3 0

3 years ago