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Lilit [14]
2 years ago
13

Consider the reaction H2(g) + Cl2(g) → 2HCl(g)ΔH = −184.6 kJ / mol If 2.00 moles of H2 react with 2.00 moles of Cl2 to form HCl,

what is ΔU (in kJ) for this reaction at 1.0 atm and 25°C? Assume the reaction goes to completion.
Chemistry
1 answer:
zalisa [80]2 years ago
4 0

Answer:

ΔU=-369.2 kJ/mol.

Explanation:

We start from the equation:

Δ(H)=ΔU+Δ(PV), which is an extension of the well known relation: H=U+PV.

If Δ(PV) were calculated by ideal gas law,

PV=nRT

Δ(PV)=RTΔn.

Where Δn is the change of moles due to the reaction; but, this reaction does not give a moles change (Four moles of HCl produced from 4 moles of reactants), so Δ(PV)=0.

So, for this case, ΔH=ΔU.

The enthalpy of reaction given is for one mole of reactant, so the enthalpy of reaction for the reaction of interest must be multiplied by two:

2 reactant moles*\frac{-184.6kJ}{mol}

ΔU=-369.2 kJ/mol.

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Methane, CH4, reacts with I2 according to the reaction CH4(g)+I2(g)⇌CH3I(g)+HI(g)
gtnhenbr [62]

Answer:

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

Explanation:

Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

Kp = \frac{(pC)^cx(pD)^d}{(pA)^ax(pB)^b}, where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr   7.96 torr  0       0            <em> initial partial pressure</em>

-x                  -x            +x     +x          <em> react</em>

105.1-x       7.96-x      x        x            <em>equilibrium</em>

Then:

Kp = \frac{pCH3IxpHI}{pCH4xpI2} = \frac{x^2}{(105.1-x)(7.96-x)}

2.26x10^{-4} = \frac{x^2}{836.596 - 113.06x -x^2}

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

0.9997x² + 0.0255x - 0.1891 = 0

Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

x = \frac{-b+/-\sqrt{0.7568} }{2a} = \frac{-0.0255 +/-0.8699}{1.9994}

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

8 0
2 years ago
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How many moles are in 1.50 g of ethylamine, ch3ch2nh2?
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Mass =  mass/molar  mass  of   ch3ch2nh2
the   molar  mass  of CH3CH2NH2 =  12  +(1x3)+12+(1 x2)+14+(1x2) =45  g/mol

moles  is therefore  =  1.50g /45g/mol =   0.033 moles
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You decide to establish a new temperature scale on which the melting point of ammonia (-77.75 ∘c) is 0∘a, and the boiling point
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-33.75-(-77.75) / 100 = 100-(-77.75) / x
44.4/100 = 177.75 / x
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The boiling point of water in ∘a would be 400.33∘a.

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Caffeine, a molecule found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 10.0 g of caffeine pr
denis-greek [22]

Answer:

194 g/mol.

Explanation:

Hello,

In this case, one first must compute the mass of each element as shown below:

C=18.13gCO_2*\frac{12gC}{44gCO_2} =4.945gC\\H=4.639gH_2O*\frac{2.016gH}{18.0152gH_2O}=0.519gH\\N=2.885gN_2\\O=10.0g-4.945g-0.519g-2.885g=1.651gO

Next, the corresponding moles:

C=4.945gC*\frac{1molC}{12gC}=0.412mol\\H=0.519gH*\frac{1molH}{1gH}=0.519mol\\N=2.885gN*\frac{1molN}{14gN}=0.206molN\\O=1.648gO*\frac{1molO}{16gO} =0.103molO

Then, each element's subscripts is found to be:

C=\frac{0.412}{0.103}=4\\H=\frac{0.519}{0.103}=5\\N=\frac{0.206}{0.103} =2\\O=\frac{0.103}{0.103}=1

Therefore, the empirical formula is:

C_4H_5N_2O

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

C_8H_10N_4O_2

Which has a molar mass of 194 g/mol being correctly contained in the given interval.

Best regards.

8 0
2 years ago
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