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Wittaler [7]
3 years ago
6

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li

st of thermodynamic properties.HCl(g)+NaOH(s)⟶NaCl(s)+H2O(l)
Chemistry
1 answer:
Anastasy [175]3 years ago
7 0

Answer:

-179.06 kJ

Explanation:

Let's consider the following balanced reaction.

HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)

We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))

ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)

ΔH°r = -179.06 kJ

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2) A common "rule of thumb" -- for many reactions around room temperature is that the
babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$  and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}} $

$\Rightarrow \frac{K_1}{K_2}=  e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

           R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}=  3.479 \times 10^{-6}$

$\Rightarrow  \frac{K_2}{K_1}=  e^{3.479 \times 10^{-6}}$

$\Rightarrow  \frac{K_2}{K_1}=  1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0
2 years ago
Someone help me please, I will give you brainliest
Nikitich [7]

Answer:

conserve mass.

Explanation:

3 0
3 years ago
Express the following in liters at STP:<br><br><br><br> 6.62 x 10-3 moles HF
serious [3.7K]

Answer:

The volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L

Explanation:

Given data:

Number of moles of HF =  6.62×10⁻³ mol

Volume of HF in litter at STP = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

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

T = temperature in kelvin

Standard temperature = 273 K

Standard pressure = 1 atm

Now we will put the values in formula.

1 atm × V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K   × 273 K

V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K   × 273 K  / 1 atm

V = 148.38×10⁻³ L

Thus, the volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L.

8 0
3 years ago
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5 0
2 years ago
A solution has a pH of 12. This solution is:
kolezko [41]

Answer:

basic

Explanation:

pH 7: neutral

12>7

so it is basic

(if <7 than acidic)

3 0
3 years ago
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