What is the total energy change for the following reaction: H2 + I2 -> 2HI?

Korolek [52]

<h2>Hello!</h2>

The answer is:

The new volume is 2.84 L.

$V_{2}=2.84L$

To solve the problem, we need to remember what STP means. STP means that the gas is at standard temperature and pressure, or 273.15 K (0°C) and 1 atm.

Also, we need to use the Combined Gas Law, since the temperature, the pressure and the volume are being changed.

The Combined Gas Law establishes a relationship between the temperature, the pressure and the volume of an ideal gas using , Gay-Lussac's Law, Charles's Law, and Boyle's Law.

The law is defined by the following equation:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}$

Where,

$P_{1}$ is the first pressure.

$V_{1}$ is the first volume.

$T_{1}$ is the first temperature.

$P_{2}$ is the second pressure.

$V_{2}$ is the second volume.

$T_{2}$ is the second temperature.

So, we are given the following information:

$V_{1}=2L\\P_{1}=1atm\\T_{1}=0\°C=273.15K\\P_{2}=80kPa=0.8atm\\T_{2}=37\°C=37+273=310K$

Then, isolating the new volume, and substituting, we have:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}\\\\V_{2}=\frac{P_{1}V{1}}{T_{1}}*\frac{T_{2}}{P_{2}}\\\\V_{2}=\frac{1atm*2L}{273.15K}*\frac{310K}{0.8kPa}=2.84L$

Hence, the new volume is 2.84 L.

$V_{2}=2.84L$

Have a nice day!

4 0

4 years ago

Describe how spead, velocity and<br> brcoloration are related.

Flura [38]

Answer:

Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance (a scalar quantity) per time ratio. On the other hand, velocity is a vector quantity; it is direction-aware. Velocity is the rate at which the position changes.

3 0

4 years ago

Describe reflection, refraction, and absorption.

Sergeeva-Olga [200]

Answer:

Refraction is the bending of a wave when it enters a medium where its speed is different, like going from the air (gas) to a glass of water (liquid). Absorption is when light energy penetrates an object. Typically this energy is then converted to heat.

7 0

3 years ago

What is the answer?

nevsk [136]

i believe that it is b

8 0

3 years ago

The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature: Time (s) (

harina [27]

Answer:

Part a: The rate of the equation for 1st order reaction is given as $Rate=k[H_2O_2]$

Part b: The integrated Rate Law is given as $[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c: The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d: Concentration after 4000 s is 0.043 M.

Explanation:

By plotting the relation between the natural log of concentration of $H_2O_2$ , the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.

Part a

Rate Law

The rate of the equation for 1st order reaction is given as

$Rate=k[H_2O_2]$

Part b

Integrated Rate Law

The integrated Rate Law is given as

$[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c

Value of the Rate Constant

Value of the rate constant is given by using the relation between 1st two observations i.e.

t1=0, M1=1.00

t2=120 s , M2=0.91

So k is calculated as

$-k(t_2-t_1)=ln{\frac{M_2}{M_1}}\\-k(120-0)=ln{\frac{0.91}{1.00}}\\k=\frac{-0.09431}{-120}\\k=7.8592 \times 10^{-4} s^{-1}$

The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d

Concentration after 4000 s is given as

$-k(t_2-t_1)=ln{\frac{M_2}{1.0}}\\-7.8592 \times 10^{-4}(4000-0)=ln{\frac{M_2}{1.00}}\\-3.1437=ln{\frac{M_2}{1.00}}\\M_2=e^{-3.1437}\\M_2=0.043 M$

Concentration after 4000 s is 0.043 M.

7 0

3 years ago