Round to four significant figures 0.00238866

Calculate the ΔG for the following system. Then state if the system is spontaneous or not spontaneous. ΔH = 147 kJ ΔS = -67.0 J/

lbvjy [14]

Answer:

ΔG for the given system is 28.421 kJ. The system is non spontaneous.

Explanation:

ΔH = 147 kJ =147,000 J= Enthalpy of the system

ΔS = -67.0 J/K = Entropy change

T = 149 °C = 422.15 K = Temperature of the system

((T)°C=273.15 +T K)

ΔG = Gibbs free energy ?

The Gibbs free energy 's expression is given by:

ΔG = ΔH - T ΔS

ΔG > 0 , non spontaneous
ΔG < 0, spontaneous

$\Delta G^o=147,000 J-422.15 K\times (-67.0)$

ΔG = 28,421 J = 28.421 kJ

Since, value of Gibbs free energy is positive this means that reaction is non spontaneous.

5 0

3 years ago

When physical properties are measured their identity will be changed. True or False

wolverine [178]

Answer:

False

Explanation:

A physical change is a change to a sample of matter in which some properties of the material change, but the identity of the matter does not.

5 0

2 years ago

1. How many molecules are in 2.40 moles of H2O?

anzhelika [568]

Answer:

There arr 1.5*1024 molecules in 2.40 moles of H2O.

4 0

3 years ago

1. A gas has a volume of 140L at 37 ºC and 620 kpa pressure. Calculate its volume at STP.

solniwko [45]

Answer:

1. Volume as STP = 755 L

2. Outside temperature = 255 K

3. Percentage yield = 70.5%

Explanation:

1. At STP, pressure = 101.3 kpa, temperature = 0°C or 273.15 K

Using the general gas equation :

P1V1/T1 = P2V2/T2

P1 = 620 kpa

V1 = 140 L

T1 = 37°C or (273.15 + 37) K = 310.15 K

P2 = 101.3 kpa

V2 = ?

T2 = 273.15 K

V2 = P1V1T2/P2T1

V2 = 620 × 140 × 273.15 / 101.3 × 310.15

V2 = 755 L

2. Using Charles' gas law:

V1/T1 = V2/T2

V1 = 2.5 L

T1 = 290 K

V2 = 2.2 L

T2 = ?

T2 = V2T1/VI

T2 = 2.2 × 290 / 2.5

T2 = 255 K

3. Equation of reaction : 2 Al + 3 CuSO4 ---> Al2 (SO4)3 + 3 Cu

From equation of the reaction, 2 moles of Al produces 3 moles of Cu

Molar mass of Al = 27 g; Molar mass of Cu = 63.5 g

2 moles of Al = 2 × 27 g = 54 g; 3 moles of Cu = 3× 63.5 = 190.5 g

54 g of Al produces 190.5 g of Cu

1.87 g of Al will produce 190.5/54 × 1.87 g of Cu = 6.60 g of Cu

Percentage yield = actual yield /theoretical yield × 100%

Percentage yield = 4.65/6.60 × 100%

Percentage yield = 70.5%

5 0

2 years ago

Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2

maxonik [38]

Answer:

Equilibrium constant of the given reaction is $1.3\times 10^{-29}$

Explanation:

$NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr$ .... $(K_{p})_{1}=5.3$

$2NO\rightleftharpoons N_{2}+O_{2}$ .... $(K_{p})_{2}=2.1\times 10^{30}$

The given reaction can be written as summation of the following reaction-

$2NO+Br_{2}\rightleftharpoons 2NOBr$

$N_{2}+O_{2}\rightleftharpoons 2NO$

......................................................................................

$N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr$

Equilibrium constant of this reaction is given as-

$\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}$

$=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})$

$=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}$

$=\frac{(5.3)^{2}}{2.1\times 10^{30}}=1.3\times 10^{-29}$

7 0

2 years ago

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