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soldi70 [24.7K]
3 years ago
12

Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the follo

wing endothermic reaction. 2N2(g) + O2(g)2N2O(g) Clear All > 0 Hrxn < 0 Srxn = 0 Grxn > 0 low T, < 0 high T Suniverse < 0 low T, > 0 high T
Chemistry
1 answer:
siniylev [52]3 years ago
3 0

Answer:

∆H > 0

∆Srxn <0

∆G >0

∆Suniverse <0

Explanation:

We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.

Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.

The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.

Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.

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1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur
marta [7]

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that P_{1} = 100\,kPa, V_{1} = 500\,mL and V_{2} = 1000\,mL, then the new pressure of the gas is:

P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)

P_{2} = 500\,kPa

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that V_{1} = 3.60\,L, P_{1} = 10\,kPa and P_{2} = 25\,kPa then the new volume of the gas is:

V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)

V_{2} = 1.44\,L

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that \frac{P_{2}}{P_{1}} = 3, then the volume ratio is:

\frac{V_{1}}{V_{2}} = 3

\frac{V_{2}}{V_{1}} = \frac{1}{3}

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

8 0
2 years ago
One of relatively few reactions that takes place directly between two solids at room temperature is <img src="https://tex.z-dn.n
kolbaska11 [484]

Answer:

a) Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)

b) 3.14g must be added

Explanation:

a) For the reaction:

Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) + H₂O(l) + NH₃(g)

As you see, there are 8 moles of water in reactants and 2 moles of oxygen in octahydrate, thus, water moles must be 10:

Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + NH₃(g)

To balance hydrogens, the other coefficients are:

Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)

b) As you see in the balanced reaction, 1 mole of barium hydroxide octahydrate reacts with 2 moles of NH₄SCN. 6.5g of Ba(OH)₂.8H₂O are:

6.5 g × (1mol / 315.48g) =<em> 0.0206moles of  Ba(OH)₂.8H₂O</em>. Thus, moles of NH₄SCN that must be used for a complete reaction are:

0.0206moles of  Ba(OH)₂.8H₂O × ( 2 mol NH₄SCN / 1 mol Ba(OH)₂.8H₂O) = <em>0.0412moles of NH₄SCN</em>. In grams:

0.0412moles of NH₄SCN × ( 76.12g / 1mol) = <em>3.14g must be added</em>

8 0
3 years ago
Read 2 more answers
A gaseous mixture consisting of nitrogen, argon, and oxygen is in a 3.5-L vessel at 25C. Determine the number of moles of oxygen
Step2247 [10]

Answer:

Number of moles of oxygen = 0.037  mol

Explanation:

Given data:

Total pressure = 98.5 KPa

Partial pressure of nitrogen = 22.0 KPa

Partial pressure of argon = 50.0 KPa

Volume = 3.5 L

Temperature = 25°C (25+273= 298K)

Number of moles of oxygen = ?

Solution:

Total pressure = P(N₂) + P(O₂) + P(Ar)

98.5 KPa = 22.0 KPa +P(O₂) + 50.0 KPa

98.5 KPa = 72.0 KPa +P(O₂)

P(O₂)  = 98.5 KPa - 72.0 KPa

P(O₂)  = 26.5 KPa

KPa to atm:

26.5 KPa/ 101 = 0.262 atm

Number of moles of oxygen:

PV = nRT

n = PV/RT

n = 0.262 atm × 3.5 L / 0.0821 atm.L/mol.K  × 298 K

n = 0.917atm.L /24.47atm.L/ mol

n = 0.037  mol

6 0
2 years ago
Consider the reaction: N2(g) 2 O2(g)N2O4(g) Write the equilibrium constant for this reaction in terms of the equilibrium constan
Valentin [98]

Answer : The equilibrium constant for this reaction is, K=\frac{(K_b)^2}{K_a}

Explanation :

The given main chemical reaction is:

N_2(g)+2O_2(g)\rightarrow N_2O_4(g);  K

The intermediate reactions are:

(1) N_2O_4(g)\rightarrow 2NO_2(g);  K_a

(2) \frac{1}{2}N_2(g)+O_2(g)\rightarrow NO_2(g);  K_b

We are reversing reaction 1 and multiplying reaction 2 by 2 and then adding both reaction, we get:

(1) 2NO_2(g)\rightarrow N_2O_4(g);  \frac{1}{K_a}

(2) N_2(g)+2O_2(g)\rightarrow 2NO_2(g);  (K_b)^2

Thus, the equilibrium constant for this reaction will be:

K=\frac{1}{K_a}\times (K_b)^2

K=\frac{(K_b)^2}{K_a}

Thus, the equilibrium constant for this reaction is, K=\frac{(K_b)^2}{K_a}

5 0
2 years ago
08.04 MC) An unknown solution has a pH of 7.2. Which of these chemicals is likely to cause the greatest decrease in the pH of th
OLEGan [10]

Answer:

HNO3

Explanation:

i have the test

6 0
3 years ago
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