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brilliants [131]

2 years ago

14

Phosphorous and chlorine gases combine to produce phosphorous trichloride: P2(g)+3Cl2(g)→2PCl3(g) ΔG∘ at 298K for this reaction

is −642.9kJ/mol. The value of ΔG at 298K for a reaction mixture that consists of 1.5atmP2, 1.6atmCl2, and 0.65atmPCl3 is ________ kJ/mol. Phosphorous and chlorine gases combine to produce phosphorous trichloride: at for this reaction is . The value of at for a reaction mixture that consists of , , and is ________ . −7.28×103 −44.2 −708.4 −3.88 ×103 −649.5

1 answer:

Wewaii [24]2 years ago

6 0

Answer:

The value of ΔG of the reaction at 298 K is -6.495 kJ/mol.

Explanation:

$P_2(g)+3Cl_2(g)\rightarrow 2PCl_3(g)$

partial pressure of phosphorus = $=p_{P_2}=1.5 atm$

partial pressure of chlorine gas= $=p_{Cl_2}=1.6 atm$

partial pressure of phosphorus trichloride = $=p_{PCl_3}=0.65 atm$

The expression of Quotient of the reaction:

$Q=\frac{[p_{PCl_3}]^2}{[p_{P_2}][p_{Cl_2}]^3}$

$Q=\frac{(0.65 atm)^2}{1.5 atm\times (1.6 atm)^3}=0.06877$

$\Delta G=\Delta G^o+-RT\ln Q$

where,

ΔG = Gibbs's free energy at 298K

ΔG° = Gibbs's free energy at equilibrium = -642.9kJ/mol = -642900 J/mol

R = Gas constant = $8.314J/K mol$

T = temperature = =298 K

$Q$ = Reaction quotient = 0.06877

Putting values in above equation, we get:

$\Delta G=-642900 J/mol+(8.314J/Kmol)\times 298K\times \ln (0.06877)$

$\Delta G=-649,532.43 J/mol=-6.495 kJ/mol$

The value of ΔG of the reaction at 298 K is -6.495 kJ/mol.

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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

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2 years ago

BRAINLIESTTT ASAP!!!

Stels [109]

While I am not the brainliest I can certainly answer.

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2 years ago

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A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m

ioda

In this kind of exercises, you should use the "ideal gas" rules: PV = nRT
P should be in Pascal:
445mmHg = 59328Pa
1225mmHg = 163319Pa

V should be in cubic meter:
16L = 0.016 m3

R = $\frac{PV}{nT}$ = constant
$\frac{P1 V1}{n T}$ = $\frac{P2 V2}{n T}$
==> P1 * V1 = P2 * V2
V2 = $\frac{P1 V1}{P2}$ = $\frac{445 0.016}{1225}$
V2 = 0.00581 m3 = 5.81 L

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2 years ago

Find the volume of a balloon of a gas at 842 mm Hg and -23 celcius if it’s volume is 915 milliliters at a pressure of 1170 mm Hg

Ronch [10]

The volume of a balloon f a gas at 842 mm Hg and -23 celsius if it’s volume is 915 milliliters at a pressure of 1170 mm Hg And a temperature of 24 celsius is 0.22 litres

Explanation:

Data given:

Initial volume of the balloon having gas V1= 915ml OR 0.195 L

initial pressure of the gas P1= 1170 mm Hg OR 1.53 atm

initial temperature of the gas T1 = 24 celsius or 273.15 + 24 = 297.15 K

Final pressure of the gas P2 = 842 mm Hg or 1.10 atm

final temperature of the gas T2 = -23 degrees or 273.15 - 23 = 250.15 K

Final volume at final temperature and pressure V2=?

The formula used is of Gas Law:

$\frac{P1V1}{T1}$ = $\frac{P2V2}{T2}$

V2 = $\frac{P1V1T2}{T1P2}$

putting the values in the equation:

V2 = $\frac{1.53 X 0.195 X 250.15}{297.15 X 1.10}$

V2 = 0.22 litres is the volume

The volume is 0.22 litres at a pressure of 1170 mmHg and temperature of -23 degrees.

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After going through a guided tutorial by selecting Run Grams Demonstration, you can create your own experiment by clicking the R

Step2247 [10]

Answer: Thus 24.0 g of $SO_2$ would be needed.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} O_2=\frac{6.00g}{32g/mol}=0.1875moles$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(l)$

According to stoichiometry :

1 mole of $O_2$ require = 2 moles of $SO_2$

Thus 0.1875 moles of $O_2$ will require= $\frac{2}{1}\times 0.1875=0.375moles$ of $SO_2$

Mass of $SO_2=moles\times {\text {Molar mass}}=0.375moles\times 64g/mol=24.0g$

Thus 24.0 g of $SO_2$ would be needed to completely react with 6.00 g of $O_2$ such that all reactants could be consumed.

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2 years ago