The rates of ordinary chemical reactions roughly double with a 10 K increase in
ID: 494352 • Letter: T
Question
The rates of ordinary chemical reactions roughly double with a 10 K increase in temperature around room temperature. On the other hand, the rates of many important biological reactions increase approximately 100-fold with a 10 K increase in temperature around room temperature. Which type of reaction has the greater activation energy? Show your reasoning. (b) If concentrations are expressed in moles per liter, and time in seconds, what are the units of (i) the rate and (ii) the specific rate constant, k, of a reaction that obeys the rate law k[A][B]^1/2 ? (c) The rate law for the forward reaction in CI_2 (g) + CO(g) rightarrow COCI_2 (g) Is found to be k[CO][CI_2]^3/2. Does this reaction occur in a single step? Explain your reasoning. (d) The rate law for the forward reaction in 3 HNO_2 (aq) doubleheadarrow H^+ (aq) + NO_3^- (aq) + 2NO(g) + H_2 O is found to be k[HNO_2]/[NO]_2. What is the rate law for the reverse reaction?Explanation / Answer
From Arhenius equation ln (K2/K1)= (Ea/R)*(1/T1-1/T2)
T1= 298, T2= 298+10=308
For ordinary reactions, ln10 =(Ea/R)* (1/298-1/308), R= 8.314*10-3 KJ/mole.K
Ea= 175.7 Kj/mole
For biochemical reaction, ln 100 = (Ea/R)*(1/298-1/308)
Ea= 351.42 Kj/mole. Biochemical reactions will have more activation energy.
2. Rate law= dCA/dt= moles/Liter.time = K[A] [B]0.5
Units of K ( the rate constant)= (mole/liter time)/ ( moles/Liter)* (moles/Liter)0.5 = 1/(Moles/Liter)0.5* time
3. The reaction does not occur in single step since there is no correspondence between the rate expression and the stoichiometry of the reaction. If the reaction had occured in single step, rate would have been = K[ CO] [Cl2], which is not the case here.
4.since it is equilibrium reaction,Rate of decomposition of HNO2/3 = rate of decomosition of NO/2=
Rate of reverse reaction = (2/3)* K[ HNO2] /[NO]2 = (2/3) times rate of forward reaction ( 2,3 are coefficients of the HNO2 and NO in the reaction )
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