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eimsori [14]
2 years ago
15

Consider the generic reaction: 2 A(g) + B(g) → 2 C(g). If a flask initially contains 1.0 atm of A and 1.0 atm of B, what is the

pressure in the flask if the reaction proceeds to completion? (Assume constant volume and temperature.)a. 1.0 atmb. 1.5 atmc. 2.0 atmd. 3.0 atm
Chemistry
1 answer:
irina1246 [14]2 years ago
3 0

Answer:

b. 1.5 atm.

Explanation:

Hello!

In this case, since the undergoing chemical reaction suggests that two moles of A react with one moles of B to produce two moles of C, for the final pressure we can write:

P=P_A+P_B+P_C

Now, if we introduce the stoichiometry, and the change in the pressure x we can write:

P=1.0-2x+1.0-x+2x

Nevertheless, since the reaction goes to completion, all A is consumed and there is a leftover of B, and that consumed A is:

x=\frac{1.0atm}{2}=0.5atm

Thus, the final pressure is:

P=1.0-2(0.5)+1.0-(0.5)+2(0.5)\\\\P=1.5atm

Therefore the answer is b. 1.5 atm.

Best regards!

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1. 51.2 g of NaClO (pKa of HClO is 7.50) and 25.3 g of KF (pKa of of HF is 3.17) were combined in enough water to make 0.250 L o
Andrews [41]

Answer:

* pH=11.0

* [HClO]_{eq}=9.32x10^{-4}M

* [HF]_{eq}=5.07x10^{-6}M

Explanation:

Hello,

At first, the molarities of NaClO and KF are computed as shown below:

M_{NaClO}=\frac{51.2g*\frac{1mol}{74.45g} }{0.250L}=2.75M \\\\M_{KF}=\frac{25.3g*\frac{1mol}{58.1g} }{0.250L} =1.74M

Now, since both NaClO and KF are ionic, one proposes the dissociation reactions as:

NaClO\rightarrow Na^++ClO^-\\KF\rightarrow K^++F^-

In such a way, by writing the formation of HClO and HF respectively, due to the action of water, we obtain:

ClO^-+H_2O\rightleftharpoons HClO+OH^-\\F^-+H_2O\rightleftharpoons HF+OH^-\\

Now, the presence of OH⁻ immediately implies that Kb for both NaClO and KF must be known as follows:

Kb=\frac{Kw}{Ka} \\Kb_{NaClO}=\frac{1x10^{-14}}{10^{-7.50}}=3.16x10^{-7}\\Kb_{KF}=\frac{1x10^{-14}}{10^{-3.17}}=1.48x10^{-11}

In such a way, the law of mass action for each case is:

Kb_{NaClO}=3.16x10^{-7}=\frac{[HClO][OH]^-}{[ClO^-]}=\frac{(x_{ClO})(x_{ClO})}{2.75-x_{ClO}}\\Kb_{KF}=1.48x10^{-11}=\frac{[HF][OH]^-}{[F^-]}=\frac{(x_{F})(x_{F})}{1.74-x_{F}}

Now, by solving for the change x for both the ClO⁻ and F⁻ equilibriums, we obtain:

x_{ClO}=9.32x10^{-4}M\\x_{F}=5.07x10^{-6}M

Such results equal the concentrations of OH⁻ in each equilibrium, thus, the total concentration of OH⁻ result:

[OH^-]_{tot}=9.32x10^{-4}M+5.07x10^{-6}M=9.37x10^{-4}M

With which the pOH is:

pOH=-log(9.37x10^{-4})=3.03

And the pH:

pH=14-pOH=11.0

In addition, the equilibrium concentrations of HClO and HF equals the change x for each equilibrium as:

[HClO]_{eq}=x_{ClO}=9.32x10^{-4}M

[HF]_{eq}=x_{F}=5.07x10^{-6}M

Best regards.

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