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3 years ago

2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)

Chemistry

1 answer:

Ray Of Light [21]3 years ago

6 0

Answer:

There will be produced 58.14 grams of Cl2; 22.97 grams of N2 and 98.45 grams of HF

Explanation:

Step 1: The balanced equation

2ClF3(g) + 2NH3(g) → N2(g) + Cl2(g) + 6HF(g)

Step 2: Data given

Mass of NH3 = 28.0 grams

Molar mass of NH3 = 17.03 grams

Mass of ClF3 = 168.8 grams

Molar mass ClF3 = 92.45 g/mol

Molar mass of N2 = 28.01 g/mol

Molar mass of Cl2 = 70.9 g/mol

Molar mass of HF = 20.01 g/mol

Step 3: Calculate moles

Moles of NH3 = mass NH3 / Molar mass NH3

Moles NH3 = 28.0 / 17.03 = 1.64 moles

Moles ClF3 = mass ClF3 / Molar mass ClF3

Moles ClF3 = 168.8 grams / 92.45 g/mol

Moles ClF3 = 1.83 moles

Step 4: Calculate limiting reactant

For 2 moles of NH3 we need 2 moles of ClF3 to produce 1 mole of Cl2, 1 mole of N2 and 6 moles of HF

NH3 is the limiting reactant. It will completely be consumed (1.64 moles).

ClF3 is in excess. There will be consumed 1.64 moles. There will remain 1.83 - 1.64 = 0.19 moles of ClF3

0.19 moles of ClF3 = 0.19 * 92.45 = 17.57 grams

Step 5: Calculate moles of Cl2, N2 and HF

For 2 moles of NH3 we need 2 moles of ClF3 to produce 1 mole of Cl2, 1 mole of N2 and 6 moles of HF

For 1.64 moles of NH3 consumed, we will produce:

1.64/ 2 = 0.82 moles of Cl2

1.64/2 = 0.82 moles of N2

1.64 * 3 = 4.92 moles of HF

Step 6: Calculate mass of products

Mass = moles * Molar mass

Mass of Cl2 = 0.82 moles * 70.9 g/mol = 58.138 grams

Mass of N2 = 0.82 moles * 28.01 g/mol = 22.97 grams

Mass of HF = 4.92 moles * 20.01 g/mol = 98.45 grams

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Answer : The initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

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:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

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

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

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

Putting values in equation 1, we get:

$6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1$

$\text{Rate}=3.4\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

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4. 264.6J

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