help with 4.11 please The rate constant for the decomposition of NO_2 is 0.255 M

Question

help with 4.11 please

The rate constant for the decomposition of NO_2 is 0.255 M^-1S^-1. Using the units of k. first determine if this is a zero-, first-, or second-order reaction. Then answer this question: If a 1.00 L reaction vessel initially contains 1.33 M NO_2, how many moles remain after 4.00 s? 1.8 mol 0.85 mol 0.48 mol 0.77 mol 0.56 mol What is the activation energy (E_a) for the reaction with the following data? 42.0 kJ/mol 23.8 kJ/mol 11.5 kJ/mol 12.5 kJ/mol 55.0 kJ/mol What is the activation energy (E_a) for a reaction with a rate constant that triples as it is heated from 293K to 308K? 42.0 kJ/mol 23.8 kJ/mol 11.5 kJ/mol 12.5 kJ/mol 55.0 kJ/mol Is the following mechanism valid for the overall reaction shown here? Overall Reaction: A+B right arrow AB, rate_obs=k[A]^2[B] Proposed Mechanism: Step one(fast) 2A equivalent A_2 Step two(slow) A_2+B right arrow AB+A No, because the elementary steps do not add up to overall reaction. No, because the deducted rate law from the elementary steps does not match the observed rate law. Yes, because the elementary steps do add up to overall reaction even though the deducted rate law from the elementary steps does not match the observed rate law. Yes, because the elementary steps do add up to overall reaction and the deducted rate law from the elementary steps does not match the observed rate law. There is not enough information provided to determine this.

Explanation / Answer

Arhenius equation for two different temperatures T1= 293 K and T2= 308 K can be written as

ln (K2/K1)= (Ea/R)*(1/T1-1/T2)

Ea= activation energy, R = gas constant = 8.314 j/mol.K

K2/K1= ratio of rate constants =3

ln3= (Ea/R)*(1/293-1/308)= (Ea/R)*0.000166

Ea =54941.75 joules= 54.94 Kj ( E is the correct answer)

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