All categories

3 years ago

How will temperature affect the spontaneity of a reaction with positive H and S?

Chemistry

1 answer:

BartSMP [9]3 years ago

5 0

Answer:

D.A high temperature will make it spontaneous.

Explanation:

we know that

Δ G = Δ H - TΔS

For spontaneous reaction

Δ G should be negative . If Δ G = 0

Δ H = TΔS

T = Δ H / Δ S

At temperature above this T , Δ G becomes negative if Δ S and ΔH are positive .

So at a high temperature will make it spontaneous.

You might be interested in

in 1990, the men's singles winner of the us. open tennis tournament has his serves clocked at 127 mi/hr. How fast must a 56.6 g

Naddika [18.5K]

This problem is providing information about the mass of a tennis ball, 56.6 g (0.0566 kg) and asks for the velocity it will have to equal the wavelength of green light, which is 5400 A or 540 nm (5.4x10⁻⁷ m). Thus, after doing the math, the result is 2.17x10⁻²⁶ m/s.

<h3>Broglie's wavelength:</h3>

In this case, we recall the formula of the Broglie's wavelength as shown below:

$\lambda =\frac{h}{mv}$

Whereas lambda is the wavelength, h is the Planck's constant, m the mass and v the speed; thus, we solve for the speed according to the question:

$v =\frac{h}{m\lambda}$

<h3>Calculations:</h3>

Then, we just plug in the numbers we were given to get the answer:

$v =\frac{6.626x10^{-34} kg*\frac{m^2}{s} }{0.056kg*5.4x10^{-7}m}\\\\v=2.17x10^{-26}m/s$

Learn more about Broglie's wavelength: brainly.com/question/5440536

3 0

2 years ago

the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

5 0

3 years ago

Read 2 more answers

What is the volume of as sample that has a mass of 40.0 g and a density of 4.30 g/mL?

kati45 [8]

Explanation:

density = mass/volume so volume = mass/density = 40/4.30

4 0

3 years ago

Calculate the number of moles in 3.01 x 10²² atoms of calcium.

Lera25 [3.4K]

Answer:

Using dimensional analysis:

3.01x1022 molecules CO2 x 1 mol CO2/6.02x1023 molecules x 44. g CO2/mole = 2.20 g CO2

Explanation:

8 0

3 years ago

A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. what is the partial pressure o

klasskru [66]

Hello!

<span>We have the following statement data:
</span>
Data:
$P_{Total} = 800 mmHg$
$P\% N_{2} = 60\%$
$P\% O_{2} = 40\%$
$P_{partial} = ? (mmHg)$

<span>As the percentage is the mole fraction multiplied by 100:

</span> $P = X_{ O_{2} }*100$

<span>The mole fraction will be the percentage divided by 100, thus:
</span><span>What is the partial pressure of oxygen in this mixture?
</span>
$X_{ O_{2} } = \frac{P}{100}$
$X_{ O_{2}} = \frac{40}{100}$
$\boxed{X_{ O_{2}} = 0.4}$

<span>To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:
</span>
In relation to $O_{2}$ :

$\frac{P O_{2} }{P_{total}} = X_O_{2}$
$\frac{P O_{2} }{800} = 0.4$
$P_O_{2} = 0.4*800$
$\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark$
<span>
Answer:
</span><span>b. 320.0 mm hg </span>

7 0

3 years ago