For the reaction H_2(g) + S(s) right arrow H_2S(g), delta H^ degree = - 20.2 kJ/

Question

For the reaction H_2(g) + S(s) right arrow H_2S(g), delta H^ degree = - 20.2 kJ/mol and delta S^ degree = +43.1 J/K mol. Which of these statements is true The reaction is only spontaneous at low temperature. The reaction is spontaneous at all temperatures. delta G^ degree becomes less favorable as temperature increased. The reaction is spontaneous only at high temperatures. The reaction is at equilibrium at 25 ^ degree C under standard conditions. Calculate deltaG ^ degree for the reaction 3NO_2(g) + H_2O(I) right arrow 2HNO_3(I) + NO(g). 8.7 kJ/mol -192 kJ/mol 192 kJ/mol -155 kJ/mol -414 kJ/mol Ozone (O_3) in the atmosphere can reaction with nitric oxide (NO): O_3(g) + NO(g) right arrow NO_2(g) + O_2(g). Calculate the delta G^ degree for this reaction at 25^ degree C. (delta H^ degree = -199kJ/mol, delta S^ degree = -4.1 kJ/mol) 1020 kJ/mol -1.42*10^3 kJ/mol -1.22*10^3 kJ/mol -198 kJ/mol 2.00*10^3 kJ/mol Hydrogen peroxide (H_2O_2) decomposes according to the equation H_2O_2(I) right arrow H_2O(I) + 1/2 O_2(g). Calculate K_p for this reaction at 25^ degree C. (delta H^ degree = -98.2 kJ/mol, delta S^ degree = 70.1 kJ/mol) 1.3*10^-21 20.9 3.46*10^17 7.5*10^20 8.6* 10^4 Complete and balance the following redox equation. What is the coefficient of H_2O when the equation is balanced using the set of smallest whoile-numbert coefficients MnO^-_4 + SO^2-_3 right arrow Mn^2+ + SO^2-_4 (acidic solution) 3 4 5OH^- when the equation is balanced using the set of smallest whoile-numbert coefficients MnO^-_4 + I^- right arrow MnO_2 + IO^-_3 (basic solution) 1 2 4 10 None of these.

Explanation / Answer

9.

5 SO3{-2} + 2 MnO4{-} + 6 H{+} = 5 SO4{-2} + 2 Mn{+2} + 3 H2O

so the coefficient of H2O is 3.

i will give the process:

firstly we will see the oxidant and reductant

Mn is reduced, from 7+ to 2+, gaining 5 electrons
S is oxidized, from 4+ to 6+, losing 2 electrons
so we give coefficients of 5 and 2 to S and Mn

2 MnO4- + 5 SO3 2- + 6 H+ ----> 2 Mn2+ + 5 SO4 2- + 3 H2O

This reaction must (though not stated above) take place in an acidic solution so we add H+ on the left and H2O on the right to balance first the oxygen, then the hydrogen.
As a final check, charge balances, the left having 6- and the right 6- as well

10.

half reactions:

MnO4- + 4 H+ + 3e-= MnO2 + 2 H2O
2 I- = I2 + 2e-

2 MnO4- + 8 H+ + 6 I- = 2 MnO2 + 4 H2O + 3 I2

add 8 OH- on the left and on the right side

2 MnO4- + 6 I- + 4 H2O = 2 MnO2 + 3 I2 + 8 OH-

coefficient of OH- = 8

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