Some questions to ask.
1. Use of the data booklet is relevant to this question
The standard enthalpy change of formation of methane is given in the equation below.
C(graphite) + 2H2 (g) -> CH4(g) standard enthalpy change of formation = -75kJ/mol
what is the enthalpy change of atomisation of graphite?
+693 +843 +1129 +2587
tried to solve using all the methods i know but cant get the ans. only same result i got was 2587, which i think is wrong anyw.
2. The decomposition of H2O2 is known to be first order reaction
2H2O2 -> 2H2O + O2
the rate constant is found to be 4.95X10^-2 min-1. if initial concentration of H2O2 is 4.0mol dm-3, what will be the concentration of H2O2 after 42 min?
2.0 mol/dm3
1.0 mol/dm3
0.5 mol/dm3
3. is SiCl4 a simple molecular compound? thanks everyone
0.25 mol/dm3
Originally posted by anpanman:Some questions to ask.
1. Use of the data booklet is relevant to this question
The standard enthalpy change of formation of methane is given in the equation below.
C(graphite) + 2H2 (g) -> CH4(g) standard enthalpy change of formation = -75kJ/mol
what is the enthalpy change of atomisation of graphite?
+693 +843 +1129 +2587
tried to solve using all the methods i know but cant get the ans. only same result i got was 2587, which i think is wrong anyw.
2. The decomposition of H2O2 is known to be first order reaction
2H2O2 -> 2H2O + O2
the rate constant is found to be 4.95X10^-2 min-1. if initial concentration of H2O2 is 4.0mol dm-3, what will be the concentration of H2O2 after 42 min?
2.0 mol/dm3
1.0 mol/dm3
0.5 mol/dm3
0.25 mol/dm3
3. is SiCl4 a simple molecular compound? thanks everyone
Q1.
Atomization enthalpy of graphite + atomization enthalpy of H2 + 4 x (bond formation enthalpy of C-H) = -75kJ/mol
Atomization enthalpy of H2 given in data booklet.
Bond formation enthalpy of C-H given in data booklet.
Make Atomization enthalpy of graphite the subject.
Q2.
From rate constant, calculate half-life.
Then apply the formula :
Molarity (desired) / Molarity (initial) = (1/2)^n
where n = no. of half-lives.