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3 years ago

P = ? in kPa

1 answer:

4 0

This problem can be solved by using ideal gas equation PV=nRT. The unit used in the numbers is already matched so you don't need to convert any of them. The calculation would be:

PV= nRT

P* 0.602 L = 0.00801 mol *311 K* (8.314 L*kPa/mol*K)
P= 20.71 L-kPa / 0.602L
P= 34.40 kPa

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The initial reaction rate for the elementary reaction 2A + B → 4C was measured as a function of temperature when the concentrati

Artist 52 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The activation energy is 124.776 $\frac{kJ}{mole}$

b) The frequency factor is 1.77 × $10^{18}$

c) The rate constant is 0.00033 $(\frac{dm^{3} }{mole} )^{2}$ $\frac{1}{s}$

Explanation:

From the question the elementary reaction for A and B is given as

2A + B → 4C

The rate equation the elementary reaction is

- $r_{A}$ = $k$ $[A]^{2}$ $[B]$

= $k[2]^{2}[1.5]$

= 6k

$k = \frac{-r_{A} }{6}$

When temperature changes, the rate constant change an this causes the rate of reaction to change as shown on the second uploaded image.

The relationship between temperature and rate constant can be deduced from these equation

$k = Aexp(-\frac{E_{a} }{RT} )$

taking ln of both sides we have

$lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}$

Considering the graph for the rate constant $ln k$ and $(\frac{1}{T} )$ the slope from the equation is $-(\frac{E_{a} }{R})$ and the intercept is $ln A$

From the given table we can generate another table using the equation above as shown on the third uploaded image

The graph of $ln k$ vs $(\frac{1}{T} )$ is shown on the fourth uploaded image

From the graph we can see that the slope is $-(\frac{E_{a} }{R} ) = - 15008$

Now we can obtain the activation energy $E_{a}$ by making it the subject in the equation also generally R which is the gas constant is $8.145 \frac{J}{kmole}$

$E_{a} = 15008 × 8,3145\frac{J}{molK}$

$= 124\frac{KJ}{mole}$

Hence the activation energy is $= 124\frac{KJ}{mole}$

b) From the graph its intercept is $ln A = 42.019$

$A = exp(42.019)$

$=1.77 × 10^{18}$

Hence the frquency factor A is $=1.77 × 10^{18}$

c) From the equation of rate constant

$lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}$

We have

$ln k = 42.019 - 15008 * (\frac{1}{300} )$

$k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}$

Hence the rate constant is $k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}$

6 0

3 years ago

Question 1 (1 point)

andrezito [222]

Answer:

True 
B. 24 hours 
Light and Dark 
A. To spin on an Axis 
True 
Nearest P0INTS 
B. 365 days 
True 
A. January 
B. July 
Summer 
B. The tilt of the Earth on its axis
True 
¿-¿
¿-¿
B. June
A. September 
B. December and March

Explanation:

Sorry yan lang alm but I hope it helps

<h2>#CARRYONLEARNING </h2><h2>#STUDYWELL </h2>

5 0

3 years ago

Calculate ΔS° for the reaction: 4Cr(s) + 3O2(g) → 2Cr2O3(s), Substance: Cr(s) O2(g) Cr2O3(s), S°(J/K⋅mol): 23.77 205.138 81.2

Mariana [72]

Answer: The value of $\Delta S^o$ for the reaction is 1051.93 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$4Cr(s)+3O_2(g)\rightarrow 2Cr_2O_3(s)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(Cr_2O_3(s))})]-[(4\times \Delta S^o_{(Cr(s))})+(3\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(Cr_2O_3(s))}=881.2J/K.mol\\\Delta S^o_{(O_2(g))}=205.13J/K.mol\\\Delta S^o_{(Cr(s))}=23.77J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (881.2))]-[(4\times (23.77))+(3\times (205.13))]\\\\\Delta S^o_{rxn}=1051.93J/K$

Hence, the value of $\Delta S^o$ for the reaction is 1051.93 J/K

5 0

4 years ago

If an element has a half life of one million years, how long will it take to be completely gone

Aleonysh [2.5K]

It should be 2 million years

3 0

3 years ago

Are the intersection of any two half-planes necessarily a half-plane

Mashutka [201]

Answer;

The answer is No,

Explanation;

-The intersection of two parallel half-planes A and B is either a half-plane (either A or B, when A and B have similar orientation) or the empty set (when A and B have opposite orientation).

-When A and B are not parallel, their intersection is a maximal open region bounded by the two lines that define A and B, respectively. In this case, the intersection always exists and it is never a half-plane.

6 0

4 years ago