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never [62]
3 years ago
15

The reaction below has an equilibrium constant kp=2.2×106 at 298 k. 2cof2(g)⇌co2(g)+cf4(g) calculate kp for the reaction below.

cof2(g)⇌12co2(g)+12cf4(g)
Chemistry
2 answers:
AnnZ [28]3 years ago
8 0

<span>I believe the correct 2nd reaction is:</span>

cof2(g)⇌1/2 co2(g)+1/2 cf4(g)

where we can see that it is exactly one-half of the original

Therefore the new Kp is:

new Kp = (old Kp)^(1/2)

new Kp = (2.2 x 10^6)^(1/2)

<span>new Kp = 1,483.24 </span>

Irina-Kira [14]3 years ago
4 0
<h3>Kp = 1.483 . 10³</h3><h3>Further explanation</h3>

The equilibrium constant is the ratio of concentration or pressure between the results of the reaction / product and the reactant with each reaction coefficient raised

The equilibrium constant is based on the concentration (Kc) in a reaction

pA + qB -----> mC + nD

\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}

While the equilibrium constant is based on partial pressure

\large {\boxed {\bold {Kp ~ = ~ \frac {[pC] ^ m [pD] ^ n} {[pA] ^ p [pB] ^ q}}}}

The value of Kp and Kc can be linked to the formula '

\large {\boxed {\bold {Kp ~ = ~ Kc. (R.T) ^ {\Delta n}}}}

R = gas constant = 0.0821 L.atm / mol.K

=n = number of product coefficients-number of reactant coefficients

There are rules in determining the Kp value for the related reaction

  • 1. If the reaction equation is reversed, the value of Kp is also reversed
  • 2. if the reaction coefficient is divided by a factor n, then the new Kp value is the square root of n from the old Kp
  • 3. if the reaction coefficient is multiplied by the factor n, then the new Kp value is the power of n from the old Kp

2COF₂ (g) ⇌CO₂(g) + CF₄ (g) Kp = 2.2 × 10⁶ at 298 K(reaction 1)

\rm COF_2(g)--->\dfrac{1}{2}CO_2(g)+\dfrac{1}{2}CF_4(g) (reaction\:2)

For the reaction 2, the reaction coefficient is divided by 2, so the new Kp value :

\rm Kp=\sqrt{2.2.10^6}\\\\Kp=1.483.10^3

<h3>Learn more</h3>

an equilibrium constant brainly.com/question/9173805

brainly.com/question/1109930

Calculate the value of the equilibrium constant, Kc

brainly.com/question/3612827

Concentration of hi at equilibrium

brainly.com/question/8962129

 

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Explanation:

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Thus 0.625 moles of O_2 will require=\frac{2}{1}\times 0.625=1.25moles of H_2

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What volume of 0.235 M H2SO4 is needed to titrate 40.0 mL of 0.0500 M Na2CO M NaCO3 solution?
Andreyy89

Answer: The volume of 0.235 M H_2SO_4 needed to titrate 40.0 mL of 0.0500 M Na_2CO_3 is 8.51 ml

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Na_2CO_3

We are given:

n_1=2\\M_1=0.235M\\V_1=?mL\\n_2=2\\M_2=0.0500M\\V_2=40.0mL

Putting values in above equation, we get:

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