Chemistry, 02.03.2020 23:03 angeleyes4u610p6np54

A 1.87 L reaction vessel, initially at 298 K, contains chlorine gas at a partial pressure of 337 mmHg and fluorine gas at a partial pressure of 820 mmHg . What is the partial pressure of excess reactant after the reaction occurs as completely as possible

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Answer from: zoeyruckel

This is an incomplete question, here is the complete question.

Chlorine gas reacts with fluorine gas to form chlorine trifluoride.

$Cl_2(g)+3F_2(g)\rightarrow 2ClF_3(g)(s)$

A 1.87 L reaction vessel, initially at 298 K, contains chlorine gas at a partial pressure of 337 mmHg and fluorine gas at a partial pressure of 820 mmHg . What is the partial pressure of excess reactant after the reaction occurs as completely as possible?

Answer : The partial pressure of excess reactant after the reaction is, 62.6 mmHg

Explanation :

First we have to calculate the moles of $Cl_2\text{ and }F_2$

Using ideal gas equation:

$n_{Cl_2}=\frac{P_{Cl_2}V}{RT}$

where,

$P_{Cl_2}$ = partial pressure of Cl_2 = 337 mmHg = 0.443 atm

R = gas constant = 0.0821 L.mmHg/mol.K

T = temperature = 298 K

V = volume = 1.87 L

$n_{Cl_2}=\frac{(0.443atm)\times (1.87L)}{(0.0821L.atm/mol.K)\times (298K)}$

$n_{Cl_2}=0.0338mol$

and,

$n_{F_2}=\frac{P_{F_2}V}{RT}$

where,

$P_{F_2}$ = partial pressure of F_2 = 820 mmHg = 1.08 atm

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = 298 K

V = volume = 1.87 L

$n_{F_2}=\frac{(1.08atm)\times (1.87L)}{(0.0821L.atm/mol.K)\times (298K)}$

$n_{F_2}=0.0825mol$

Now we have to calculate the limiting and excess reagent,

The balanced chemical reaction is,

$Cl_2(g)+3F_2(g)\rightarrow 2ClF_3(g)(s)$

From the balanced reaction we conclude that

As, 3 mole of F_2 react with 1 mole of Cl_2

So, 0.0825 moles of F_2 react with $\frac{0.0825}{3}=0.0275$ moles of Cl_2

From this we conclude that, Cl_2 is an excess reagent because the given moles are greater than the required moles and F_2 is a limiting reagent and it limits the formation of product.

Renaming moles of Cl_2 = 0.0338 - 0.0275 = 0.00630 mol

Now we have to calculate the partial pressure of excess reactant after the reaction.

$P_{Cl_2}=\frac{n_{Cl_2}RT}{V}$

$P_{Cl_2}=\frac{(0.00630mol)\times (0.0821L.atm/mol.K)\times (298K)}{1.87L}$

$P_{Cl_2}=0.0824atm=0.0824\times 760mmHg=62.6mmHg$

Thus, the partial pressure of excess reactant after the reaction is, 62.6 mmHg

Answer from: Quest

c- cold periods in earth's history during which glaciers covered much of the surface.

explanation:

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

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A 1.87 L reaction vessel, initially at 298 K, contains chlorine gas at a partial pressure of 337 mmH...