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2 years ago

Write an equation that expresses the law of thermodynamics in terms of heat and work

Chemistry

1 answer:

Anettt [7]2 years ago

3 0

We write DE = q+w, where DE is the internal energy change and q and w are heat and work, respectively.

(b)Under what conditions will the quantities q and w be negative numbers?

q is negative when heat flows from the system to the surroundings, and w is negative when the system does work on the surroundings.

As an aside: In applying the first law, do we need to measure the internal energy of a system? Explain.

The absolute internal energy of a system cannot be measured, at least in any practical sense. The internal energy encompasses the kinetic energy of all moving particles in the system, including subatomic particles, as well as the electrostatic potential energies between all these particles. We can measure the change in internal energy (DE) as the result of a chemical or physical change, but we cannot determine the absolute internal energy of either the initial or the final state. The first law allows us to calculate the change in internal energy during a transformation by calculating the heat and work exchanged between the system and its surroundings.

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What is the range for the following set of scores? Scores:6, 12, 9, 17, 11, 4, 14

MrRissso [65]

This is the answer is 17

5 0

2 years ago

The hydrogen-line emission spectrum includes a line at a wavelength of 434 nm. What is the energy of this radiation? (h= 6.626 x

Andrews [41]

Wavelength = 434nm = 434 x 10⁻⁹m
planck's constant = h= 6.626 x 10 ⁻³⁴ J
E =?
by using the formula;
E = hc /λ
value for c is 3 x 10⁸ m/s
E = (6.626 x 10 ⁻³⁴ J)(3 x 10⁸ m/s) / 434 x 10⁻⁹m
E = 1.9878 x 10⁻²⁵ / 434 x 10⁻⁹m
E = 4.58 x 10⁻¹⁹ joules

6 0

2 years ago

If 335 g water at 35.5C loses 5750 J of heat, what is the final temperature of the water?

White raven [17]

Answer:

The final temperature is 39.58 degree Celsius

Explanation:

As we know

Q = m * c * change in temperature

Specific heat of water (c) = 4.2 joules per gram per Celsius degree

Substituting the given values we get -

5750 = 335 * 4.2 * (X - 35.5)

X = 39.58 degree Celsius

3 0

2 years ago

Reaction rate is expressed in terms of changes in the concentration of reactants and products. Write a balanced equation for the

KengaRu [80]

Answer : The balanced equations will be:

$CH_4+2O_2\rightarrow 2H_2O+CO_2$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the balanced equations corresponding to the following rate expressions.

$Rate=-\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2O]}{dt}=+\frac{d[CO_2]}{dt}$

The balanced equations will be:

$CH_4+2O_2\rightarrow 2H_2O+CO_2$

5 0

2 years ago

An industrial manufacturer wants to convert 175 kg of methane into HCN. Calculate the masses of ammonia and molecular oxygen req

Tema [17]

Given:

175 kilograms of Methane (CH4) to be synthesized into Hydrogen Cyanide (HCN)

The balanced chemical equation is shown below:

2 CH4 + 2 NH3 + 3 O2 → 2 HCN + 6 H2O

To calculate for the masses of ammonia and oxygen needed, our basis will be 175 kg CH4.

Molar mass:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol

mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol
mass of NH3 = 185.94 kg NH3 needed

mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
mass of O2 = 525 kg

mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
mass of O = 131.25 kg O

4 0

2 years ago