Review the subsection in this chapter entitled Making a Nonspontaneous Process S

Question

Review the subsection in this chapter entitled Making a Nonspontaneous Process Spontaneous in Section 18.8. The hydrolysis of ATP, ATP(aq)+H2O(l)?ADP(aq)+Pi(aq) ?G?rxn=?30.5kJ, is often used to drive nonspontaneous processes?such as muscle contraction and protein synthesis?in living organisms. The nonspontaneous process to be driven must be coupled to the ATP hydrolysis reaction. For example, suppose the nonspontaneous process is A+B?AB (?G? positive). The coupling of a nonspontaneous reaction such as this one to the hydrolysis of ATP is often accomplished by the following mechanism. A+ATP+H2OA?Pi+BA+B+ATP+H2O???A?Pi+ADPAB+PiAB+ADP+Pi As long as ?G?rxn for the nonspontaneous reaction is less than 30.5 kJ, the reaction can be made spontaneous by coupling in this way to the hydrolysis of ATP. Suppose that ATP is to drive the reaction between bicarbonate and ammonia to form carbamate. There is a scheme of a reaction, where bicarbonate reacts with ammonia to form carbamate and water. Bicarbonate has the following structure: HOCO minus, with an oxygen atom attached to the carbon atom by a double bond. Ammonia is NH3. Water is H2O. Carbamate has the following structure: H2NCO minus, with an oxygen atom attached to the carbon atom by a double bond. Part A Calculate K for the reaction between bicarbonate and ammonia. (The standard free energy change for the reaction is +3.30 kJ/mol. Assume a temperature of 298 K.) Express your answer using three significant figures. K = Request Answer Part B The bicarbonate and ammonia reaction can couple with the hydrolysis of ATP (such as shown above). What is ?G?rxn for this coupled reaction? Express your answer using three significant figures. ?G?rxn = kJ Request Answer Part C What is K for this coupled reaction? Express your answer using three significant figures. K =

Explanation / Answer

(a) Value of Keq is asked for the reaction between bicarbonate and ammonia

We know that Standard Free Energy and Keq are related by the following expression-

delta G = -RT *ln Keq or deltaG = -2.303*R*T*log K.........(I),

here it is given deltaG = +3.30 kJ/mol, T = 298 K, and R we know = 8.314*10-3 kJ/mol-K.

Now we put in the values to find Keq,

3.30 kJ/mol = -2.303* 8.314*10-3 kJ/mol-K *298 K*log Keq => 3.30 = -5.705*log Keq => log Keq = -0.578

Therefore Keq = 10^-0.578 = 0.264

(b) The Standard Free Energy deltaGrxn of the coupled reaction will be the sum of two Standard Free Energy values i.e. deltaG of ATP hydrolysis and deltaG of bicarbonate-ammonia reaction.

Therefore, deltaGrxn = -30.5 kJ + 3.30 kJ = -27.2 kJ.

(c) Keq for the coupled reaction can be calculated like in part (a), however, the value of deltaG here is equal to -27.2kJ. Now we will put all the values in eqn (I) to find Keq.

-27.2 kJ/mol = -2.303* 8.314*10-3 kJ/mol-K *298 K*log Keq => 27.2 = 5.705*log Keq => log Keq = 4.768

Therefore Keq = 10^4.768 = 58,613.816 = 58.6 x 103

Thanks!

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