True/False. A program is an algorithm that has been coded into something that can be run by a machine. *

Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g) Part A Calculate ΔG for this reaction at 25 ∘C under the following condit

kati45 [8]

Answer: The $\Delta G$ of the reaction at given temperature is -12.964 kJ/mol.

Explanation:

For the given chemical reaction:

$CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)$

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{CO})\times (p_{H_2}^2)}{p_{CH_3OH}}$

We are given:

$p_{CO}=0.140atm\\p_{H_2}=0.180atm\\p_{CH_3OH}=0.850atm$

Putting values in above equation, we get:

$K_p=\frac{(0.140)\times (0.180)^2}{0.850}\\\\K_p=5.34\times 10^{-3}$

To calculate the Gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = $5.34\times 10^{-3}$

Putting values in above equation, we get:

$\Delta G=0+(8.314J/K.mol\times 298K\times \ln(5.34\times 10^{-3}))\\\\\Delta G=-12963.96J/mol=-12.964kJ/mol$

Hence, the $\Delta G$ of the reaction at given temperature is -12.964 kJ/mol.

5 0

3 years ago

What two factors most affect the pitch of the sounds made by a person's vocal cords? A. the volume and health of the person's lu

dusya [7]

Answer:

The pitch of the sound a person makes is determined by several factors, including the size and tension of his or her vocal cords.

Explanation:

6 0

2 years ago

Rank the following 0.100 M solutions in order of increasing H3O+ concentration:

Inessa05 [86]

Answer:

HCN < HOCl < HF

Explanation:

The larger the Kₐ value, the stronger the acid.

6.2 × 10⁻¹⁰ < 4.0 × 10⁻⁸ < 6.3 × 10⁻⁴

HCN < HOCl < HF

weakest stronger strongest

5 0

3 years ago

A large balloon is initially filled to a volume of 25.0 L at 353 K and a pressure of 2575 mm Hg. What volume of gas will the bal

bija089 [108]

Answer:

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 25.0 L

V₂ = ?

P₁ = 2575 mm Hg

Also, P (atm) = P (mm Hg) / 760

P₁ = 2575 / 760 atm = 3.39 atm

P₂ = 1.35 atm

T₁ = 353 K

T₂ = 253 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac{{3.39}\times {25.0}}{353}=\frac{{1.35}\times {V_2}}{253}$

$\frac{1.35V_2}{253}=\frac{3.39\times \:25}{353}$

Solving for V₂ , we get:

V₂ = 45.0 L

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

4 0

3 years ago

A horizontal pipe 15 cm in diameter and 4 m long is buried in the earth at a depth of 20 cm. The pipe-wall temperature is 75 deg

Basile [38]

Explanation:

The given data for case (1) is as follows.

h = 20 cm = 0.2 m

Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.

Q = $kA \frac{\Delta T}{L}$

where, A = area

L = length

k = thermal conductivity = 0.8 W/m

$\Delta T$ = change in temperature.

Therefore, putting the given values into the above formula as follows.

Q = $kA \frac{\Delta T}{L}$

= $0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{0.2}$

= 168 W

For case (2), h = 180 cm = 1.8 m

Therefore, heat lost will be calculated as follows.

Q = $kA \frac{\Delta T}{L}$

= $0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{1.8}$

= 18.67 W

Thus, we can conclude that 18.67 W heat lost if the pipe was buried at a depth of 180 cm.

8 0

3 years ago