EX3 Question 6 Given the following equation, N 2 O (g) + NO 2(g) 3 NO (g) G° rxn

Question

EX3

Question 6

Given the following equation,     N2O(g) + NO2(g)   3 NO(g)        

G°rxn = -23.0 kJ

Calculate G°rxn for the following reaction:   3 NO(g) N2O(g) + NO2(g)   

·       Question 7

Calculate the pH of a buffer which contains 0.0400 M pyruvic acid and 0.0445 M sodium pyruvate. (The Ka of pyruvic acid is 4.1 x 10-3)

·       Question 9

A potassium hydroxide (KOH) solution has a concentration of 0.150M. Calculate the pH of this solution. (T = 25oC)  

Given the following equation,     N2O(g) + NO2(g)   3 NO(g)        

G°rxn = -23.0 kJ

Calculate G°rxn for the following reaction:   3 NO(g) N2O(g) + NO2(g)   

Explanation / Answer

6)

N2O(g) + NO2(g)   3 NO(g)        

G°rxn = -23.0 kJ

when we reverse the reaction, sign of G°rxn also reverses

3 NO(g) —> N2O(g) + NO2(g

so,

G°rxn = 23.0 kJ

Answer: 23.0 KJ

7)

Ka = 4.1*10^-3

pKa = - log (Ka)

= - log(4.1*10^-3)

= 2.39

use:

pH = pKa + log {[conjugate base]/[acid]}

= 2.39+ log {0.445/0.0400}

= 3.43

Answer: 3.43

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