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algol [13]
3 years ago
11

At a given temperature, 2.6 atm of h2 and 3.14 atm of cl2 are mixed and allowed to come to equilibrium. the equilibrium pressure

of hcl is found to be 1.13 atm. calculate kp for the reaction at this temperature.
Chemistry
1 answer:
Keith_Richards [23]3 years ago
7 0

The solution would be like this for this specific problem:

<span>Given:

H2 = </span><span>2.6 atm
CL2 = 3.14 atm</span>

 

<span>
pressure H2 = 2.6 - x 
pressure Cl2 = 3.14 - x 
<span>pressure HBr = 2x = 1.13

x = 1.13 / 2 = 0.565 

<span>pressure H2 = 2.6 - 0.565 = 2.035
pressure Br2 = 3.14 - 0.565 = 2.575 

Kp = (1.13)^2 / 2.035 x 2.575</span></span></span>

 

= 1.2769 / (5.240125)

= 0.24367739319195629875241525726963

= 0.244

<span>Therefore, the Kp for the reaction at the given temperature is 0.244.

To add, </span>the hypothetical pressure of a gas if it alone occupied the whole volume of the original mixture at the same temperature is called the partial pressure or Kp.

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Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2}  =0.75molCd(OH)_2

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m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2

Best regards.

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