Find the Delta H for the reaction below, given the following reactions and subse

Question

Find the Delta H for the reaction below, given the following reactions and subsequent Delta H values: N_2H_4(l) + CH_4O(l) rightarrow CH_2O(g) + N_2(g) + 3H_2 (g) 2NH_3(g) rightarrow N_2H_4(1) + H_2(g) Delta H = 22.5 kJ 2NH _3(g) rightarrow N_2(g) + 3H-2(g) Delta H = 57.5 kJ CH_2O(g) + H_2(g) rightarrow CH_4O(1) Delta H = 81.2 kJ Find the Delta H for the reaction below, given the following reactions and subsequent Delta H values: l/2H_2(g) + l/2CI_2(g) rightarrow HCl(g) COCI_2(g) + H_2O(1) rightarrow CH_2CI_2(1) + O_2(g) Delta H = 47.5 kJ 2HCI(g) + l/2O_2(g) rightarrow H_2 O(1) + CI_2(g) Delta H = 105 kJ CH_2CI_2(1) + H_2(g) + 3/2O_2(g) rightarrow COCI_2(g) + 2H_2O(1) Delta H = - 402.5 kJ Find the Delta H for the reaction below, given the following reactions and subsequent Delta H values: C_2H_2(g) + 5/2O_2(g) rightarrow 2CO_2(g) + H_2O(g) C_2H_6(g) rightarrow C_2H_2(g) + 2H_2(g) Delta H = 28J.5 kJ H_2(g) + l/2O_2(g) rightarrow H_2O(g) Delta H = - 213.7 kJ 2CO_2(g) + 3H_2O(g) rightarrow C_2H_6(g) + 7/2O_2(g) Delta H = 849 kJ Find the Delta H for the reaction below, given the following reactions and subsequent Delta H values: HCI(g) + NaNO_2(s) rightarrow HNO_2(l) + NaCI(s) 2NaCI(s) + H_2O(l) rightarrow 2HCI(g) + Na_2O(s) Delta H = 507 kj NO(g) + NO_2(g) + Na_2O(s) rightarrow 2NaNO_2(s) Delta H = - 427 kJ NO(g) + NO_2(g) rightarrow N_2O(g) + O_2(g) Delta H = - 43 kJ 2HNO_2(1) rightarrow N_2O(g) + O_2(g) + H_2O(1) Delta H = 34 kJ

Explanation / Answer

SOLUTION:

Q8. N2H4(l) + CH4O(l) --------> CH2O(g) + N2(g) + 3H2(g)

CONCEPT: Perform some algebric operations on the given other three chemical equation in such a way that we get above equation.

2NH3(g) ------> N2H4(i) + H2(g)   H = 22.5KJ =====1)

2NH3(g) ------> N2(g) + 3H2(g) H = 57.5KJ =====2)

CH2O + H2(g) ------> CH4O(l) H = 81.2KJ =====3)

Subtracting equation 1) and 3) from 2) (When we subtract a chemical equation from other, we simply reverse it and change sign of H. Then cancell out similar species that are on opposite sides)

N2H4(l) + H2(g) ------> 2NH3(g)   H = - 22.5KJ =====1)

2NH3(g) ------> N2(g) + 3H2(g) H = 57.5KJ =====2)

CH4O(l) ------>CH2O(g) + H2(g) H = - 81.2KJ =====3)

===================================================

N2H4(l) + CH4O(l) --------> CH2O(g) + N2(g) + 3H2(g) Over all reaction

H = - 22.5KJ + 57.5KJ - 81.2KJ = 46.2KJ

Q9. Procedure is similar to above. Here we will first devide the equation 2) by 2. we have

Devide all the equation by 2 as , we have

1/2COCl2 + 1/2H2O ----------> 1/2CH2Cl2 + 1/2O2   H = 47.5/2 = 23.75KJ

HCl + 1/4O2 --------> 1/2H2O + 1/2Cl2   H = 105/2 = 52.5KJ

1/2CH2Cl2 + 1/2H2 + 3/4O2 ---------->1/2COCl2 + H2O H = -402.5/2 = - 201.25KJ

Subtract 2) from 1) and 3) we have

1/2H2 + 1/2Cl2 -----------> HCl

H = 23.75KJ - 52.5KJ - 201.25KJ = - 230KJ

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