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3 years ago

5

Many manufacturing processes involve chemical reactions that reach equilibrium. Why would chemists adjust conditions to favor th

e forward reaction in a process

2 answers:

ivann1987 [24]3 years ago

7 0

To increase the amount of product

storchak [24]3 years ago

4 0

The answer is to increase the amount of product -gradpoint

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Answer:

The Sigma (σ) Bond

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S-P Overlapping. ...

P-P overlapping.

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How do you balance AlC3+H2O➡️Al(OH)3+CH4?

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4 years ago

What was Ernest Rutherford's experiment?

Dafna1 [17]

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4 years ago

What is the ideal gas law

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Explanation:

7 0

3 years ago

Read 2 more answers

You want to determine ΔH o for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) To do so, you first determine the heat capacity

Assoli18 [71]

Answer:

(A) The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

(B) ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

Explanation:

Solution

Calculate the heat actually evolved.

q = mcΔt

Finding the mass of the reactants in grams we have.

Use density. (50 mL + 50 mL ) = 100 mL of solution.

100 mL X 1.04g/mL = 104 grams of solution. (mass = Volume X Density)

Find the temperature change.

Δt =tfinal - tinitial = 30.4°C – 16.9°C = 13.5°C

q = mcΔt

= 104grams × 3.93J/g°C × 13.5°C = 5.51772×103J

= 5.51772 × 103 J

This is the heat lost in the reaction between HCl and NaOH, therefore q = -5.52 × 103 J.

this is an exothermic heat producing reaction.

To calculate the total heat of the reaction or heat per mole we have

50.0 mL of HCl X 2.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

The same quantity of base, 0.100 mole NaOH, was used.

The energy per unit mole is given by

i.e. molar enthalpy = J/mol = -5.52 × 103J / 0.100 mol

= -5.52 × 104 J/mol

= -55177.2 J/mol

= -55.177 kJ/mol

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -55.177 kJ/mol

Heat absorbed by the calorimeter = −57.32kJ − 55.177 kJ = −2.1428KJ

The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

(B) For the ZnCl we have

Calculate the heat actually evolved.

q = mcΔt

Finding the mass of the reactants in grams we have.

Use density. 100 mL of solution of HCl

100 mL X 1.015g/mL = 101.5 grams of solution. (mass = Volume X Density)

Find the temperature change.

Δt =tfinal - tinitial = 20.5°C – 16.8°C = 3.7 °C

q = mcΔt

= 101.5grams × 3.95J/g°C × 3.7°C = 1483.422×103J

= -1483.422×103J

This is the heat lost in the reaction between HCl and NaOH, therefore q = -1.483 × 103 J.

this is an exothermic heat producing reaction.

To calculate the total heat of the reaction or heat per mole we have

100.0 mL of HCl X 1.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

The energy per unit mole is given by

i.e. molar enthalpy = J/mol = -1.483 × 103J / 0.100 mol

= -1.483 × 104 J/mol

= -14834.22 J/mol

= -14.834 kJ/mol

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -14.834 kJ/mol

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

= -14.834 kJ –(0.1587KJ/°C×3.7°C) = -15.42KJ

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

5 0

3 years ago