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victus00 [196]
3 years ago
8

How to calculate delta S surroundings? calculate Delta S(surr) at the indicated temperature for a reaction having each of the fo

llowing changes in enthalpy. the first problem is delta H rxn = -283 kJ; 298 K. can anyone walk me through this so i can do all of them? how do i calculate Delta S(sys)?
Chemistry
1 answer:
astraxan [27]3 years ago
8 0
You can use the equation ΔS(surr)=q(surr)/T or ΔS(surr)=-q(rxn)/T.
the two equations are equal since we know that the energy the system (reactoin) puts out just goes into the surroundings.  
(In other words q(surr)=-q(rxn))

Using the equation, <span>ΔS(surr)=-(-283kJ/298K)=0.9497kJ/K or 949.7J/K

This answer makes sense since the reaction is exothermic which means it released energy into the system which usually causes the entropy to increase.

I hope that helps.</span>
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Many substances can go through physical and chemical changes. Which of the following is an example of a physical change? Questio
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Mixing baking soda with vinegar to form CO2

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3 years ago
50 POINTS! PLEASE HELP!A gas in a balloon at constant pressure has a volume of 120.0mL at -12.30C. What is its volume at 197.00C
Cerrena [4.2K]

Answer:

Final volume=V₂ = 216.3 mL

Explanation:

Given data:

Initial volume = 120.0 mL

Initial temperature = -12.3 °C (-12.3 +273 = 260.7 K)

Final volume = ?

Final temperature = 197.0 °C (197+273 = 470 K)

Solution:

We will apply Charles Law to solve the problem.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 120 mL × 470 K /260.7K

V₂ = 56400 mL.K /260.7K

V₂ = 216.3 mL

4 0
3 years ago
In unicorns, a heterozygous would have a rainbow tail
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2 years ago
Why is it important to clean and care for cuts on your skin
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6 0
3 years ago
A student ran the following reaction in the laboratory at 671 K: 2NH3(g) N2(g) + 3H2(g) When she introduced 7.33×10-2 moles of N
vaieri [72.5K]

Answer:

Kc = 8.05x10⁻³

Explanation:

This is the equilibrium:

                 2NH₃(g)   ⇄     N₂(g)     +     3H₂(g)

Initially       0.0733

React         0.0733α          α/2                3/2α

Eq     0.0733 - 0.0733α    α/2                0.103

We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.

Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.

3/2α = 0.103

α = 0.103 . 2/3 ⇒ 0.0686

So, concentration in equilibrium are

NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682

N₂ = 0.0686/2 = 0.0343

So this moles, are in a volume of 1L, so they are molar concentrations.

Let's make Kc expression:

Kc= [N₂] . [H₂]³ / [NH₃]²

Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³

3 0
2 years ago
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