Calculate the enthalpy of the reaction shown below, using the partial reactions

Question

Calculate the enthalpy of the reaction shown below, using the partial reactions displayed. C_6H_6(l) + (15/2)O_2(g) rightarrow 6CO_2(g) + 3H_2O(g) delta H = -3271 kJ C(s) + O_2(g) rightarrow CO_2(g) delta H = -394 kJ H_2(g) + (1/2)O_2(g) rightarrow H_2O(g) delta H = -286 kJ 6C(s) + 3H_2(g) rightarrow C_6H_6(I) delta H = ? Consider the following molecules: HF, NH_3, CHF_3 and CH_3-CH_2-CH_2-CH_2-CH_3 (Pentane) Answer the following questions: Draw the Lewis structures of the given compounds. Identify the most important intermolecular force for each of the four molecules given. Make use of delta-charges where applicable. Would hydrogen fluoride HF mix better with NH_3 or pentane? Explain. Rate data were obtained for following reaction: A + 2B rightarrow C + 2D Determine the rate law and calculate the rate constant. Calculate pH of the following solutions. Ka = 1.8 times 10^-5 for acetic acid 0.3M HCl 0.3 M CH_3COOH 0.3 M CH_3COONa

Explanation / Answer

Solution :-

a)0.3 M HCl

HCl is the strong acid therefore it dissociates completely

Therefore concentration of the H3O+ is 0.3 M

pH= -log [H3O+]

pH= -log [0.3]

pH= 0.52

b) 0.3 M CH3COOH

CH3COOH + H2O ------ > H3O^+   + CH3COO^-

0.3                                           0                    0

-x                                             +x                 +x

0.3-x                                        x                   x

Ka=[H3O^+][CH3COO^-]/[CH3COOH]

1.8*10^-5 = [x][x]/[0.3-x]

Since ka is small therefore x from the denominator can be neglected

1.8*10^-5 = [x][x]/[0.3]

1.8*10^-5 * 0.3 = x^2

5.4*10^-6 = x^2

Taking square root of both sides we get

2.32*10^-3 = x = [H3O+]

pH=- log [H3O+]

pH= -log [2.32*10^-3]

pH= 2.63

c) 0.3 M CH3COONa

it is conjugate base of acetic acid so it forms OH- when added to water

CH3COONa + H2O --------- > CH3COOH + OH^-

0.3                                           0                    0

-x                                             +x                 +x

0.3-x                                        x                   x

Kb= Kw / Ka

Kb = 1*10^-14 / 1.8*10^-5

Kb = 5.56*10^-10

Kb = [CH3COOH][OH-]/[CH3COONa]

5.56*10^-10 = [x][x]/[0.3-x]

Since kb is very small we can neglect the x from denominator

5.56*10^-10 = [x][x]/[0.3]

5.56*10^-10 *0.3 = x^2

1.67*10^-10 = x^2

Taking square root of both sides we get

1.29*10^-5 = x= [OH-]

pOH = -log [OH-]

pOH= -log [1.29*10^-5]

pOH = 4.89

pH= 14 – pOH

pH= 14 – 4.89

pH= 9.11

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