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Diano4ka-milaya [45]
3 years ago
14

5.00 mol of ammonia are introduced into a 5.00 L reactor vessel in which it partially dissociates at high temperatures. 2NH 3(g)

3H 2(g) + N 2(g) At equilibrium and a particular temperature, 1.00 mole of ammonia remains. Calculate K c for the reaction.
Chemistry
1 answer:
allochka39001 [22]3 years ago
4 0

Explanation:

system at equilibrium, will the reaction shift towards reactants ~

--?'

2. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). The production of ammonia is an

exothermic reaction. Will heating the equilibrium system increase o~e amount of

ammonia produced? . .co:(

3. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). Ifwe use a catalyst, which way will

the reaction shift? ':'\

.1.+- w~t s~,H (o')l r'eo.c. e~ ei~i"liht-,·u.fn\ P~~,

4. (3 Pts) ff 1ven th e o £ 11 owmg d t a a £ or th ere action: A(g) + 2B(s) =; AB2(g)

Temperature (K) Kc

300 1.5x104

600 55 k ' pr, cl l<..J~

e- ~ r fee, ct o. ~ 1<

900 3.4 X 10-3

Is the reaction endothermic or exothermic (explain your answer)?

t d- IS o.,;r-. \4\a..i~1f't~ °the te.Y'il(lf1,:J'u.r-a a•~S. j lrvdu..c,,.) +~H~to{' '\

exothe-rnh't.-- ,.. ..,. (/.., ,~.

5. (4 Pts) Consider the reaction, N2(g) + 3H2(g) =; 2NH3(g). Kc= 4.2 at 600 K.

What is the value of Kc for 4 NH3(g) =; 2N2(g) + 6H2(g)

N ... ~l + 3 H~(ri ~ ~Nli3~) kl,= ~:s;H,J3 # 4. J..

~ ;)N~~) ~ ~ H ~) ~\-_ == [A!;J:t D~~Jb

J. [,v 1+3] ~

I

4,:i.~ = 0,05

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A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M



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The right line is a 90° clockwise rotation of the left line about the origin. Click the 90° clockwise button. Are these lines th
inessss [21]

Answer:

Switch the coordinates and change the sign of the second one by multiplying it by negative 1.

Explanation:

Here are some examples and a more general way to understand the problem.

Consider the point (1,1), a 90 degree rotation clockwise about the origin would move it into the 4th quadrant.

The new point is (1,-1) , similarly (-4,2)-> (2,4), (-4,3)-> (3,4)

We take a point p= (x,y) the the result of rotation p 90 clockwise about the orgin is a new point p'=(x',y')= (-y, x). .

In the case of p=(1,0) the new point is p'= (0, -1)

One can use a matrix where the first row is cos(a), sin(a) and the second row is

-sin(a) cos(a) for any clockwise rotation of a degrees about the origin.

If we let a=90 degrees we have

[0 1] as the first row and [-1 0] as the second row. So the matrix is:

|0 1|

|-1 0|

Call that matrix M

So a point p= (x,y) can be multiplied by M as follows Mp=p' where p' is the rotated point.

If p=(-4,2) then Mp

is M(-4,2) which after matrix multiplication means x'=0*-4+1*2=2 and y'=-1*-4+0*2=4

So p'=(2,4)

Try it with (1,0)

x'=1*0+0*1=0

y'=-1*1+0*1=-1

so p'=(0,-1) and (1,0)->(0,-1)

How about the point on the y axis (0,1), it should go to the point (1,0)

0*1+1*1=1 and -1*0+0*1 gives you the pont (1,0) ( we don't see the negative sign because -0 is just 0)

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Usually, a model is a depiction of a certain entity, never the real thing.  In times

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