Consider these reactions, where M represents a generic metal.?

Question

Consider the following reactions, where M represents a generic metal:

( 2M(s) + 6HCl(aq) ightarrow 2MCl_3(aq) + 3H_2(g) )  (\Delta H_1 = -885.0 \, 	ext{kJ})

( HCl(g) ightarrow HCl(aq) )  (\Delta H_2 = -74.8 \, 	ext{kJ})

( H_2(g) + Cl_2(g) ightarrow HCl(g) )  (\Delta H_3 = -1845.0 \, 	ext{kJ})

( MCl_3(aq) ightarrow MCl_3(s) )  (\Delta H_4 = -161.0 \, 	ext{kJ})

Using the information above, determine the enthalpy change ((\Delta H)) for the following reaction:
( 2M(s) + 3Cl_2(g) ightarrow 2MCl_3(s) )  (\Delta H = \, __________\, 	ext{kJ})
I attempted to calculate this by using the following combination of reactions: 
1 * Reaction 1 + 2 * Reaction 4 - 6 * Reaction 2 - 3 * Reaction 3. 
However, my answer was marked incorrect, and I am unsure where I went wrong.

Golden Hansen III · Accepted Answer

friend u gone wrong in subtracting the reaction there is no need of subtraction here
just add the 1st reaction + 6 times the second reaction + 3 times the third reaction +2 times the fourth one

giving  the answer to be -7190.85kJ
hess law

Anonymous · Answer

Since 2M(s) is in the reactants of the goal equation, we know we’ll use equation #1 in the forward direction.

Since 2MCl3(s) is the products of the goal equation, we know we’ll use equation #4 in the reverse direction, and multiplied by a factor of 2.

To get 6HCl(g) and cancel out 6HCl(aq), we need to use reaction #2 multiplied by a factor of 6.

Equation #3 is useful for cancelling out HCl(g) and H2(g), but we need to multiply it by a factor of 3.

Than add up equation 1 thru three in positive number and subtract the four equation for you answer.

Anonymous

Consider these reactions, where M represents a generic metal.?

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