A total of 2.00 \rm mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 115 g of solution. The reaction caused the temperature of the solution to rise from 21.0 to 24.7 ^\circ \rm C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water.
heat gained= 115*specificeatwater*(24.7-21)
enthalpy= - heat gained by water/molescompound.
To find the enthalpy of the reaction, we need to use the formula:
q = mcΔT
Where:
q is the amount of heat transferred in Joules (J)
m is the mass of the solution in grams (g)
c is the specific heat capacity of the solution in J/g°C
ΔT is the change in temperature in °C
First, let's calculate the mass of the solution. We know that the reaction produced 115 g of solution. Therefore, the mass of the solution (m) is 115 g.
Next, we need to calculate the change in temperature (ΔT). The initial temperature (T1) was 21.0°C, and the final temperature (T2) was 24.7°C. Therefore, the change in temperature (ΔT) is:
ΔT = T2 - T1 = 24.7°C - 21.0°C = 3.7°C
Now, we need to determine the specific heat capacity (c) of the solution. We are assuming the specific heat capacity of the solution is the same as pure water. The specific heat capacity of pure water is approximately 4.18 J/g°C.
Finally, we can calculate the amount of heat transferred (q) using the formula:
q = mcΔT
q = (115 g)(4.18 J/g°C)(3.7°C)
Calculating this expression will give us the amount of heat transferred.
After calculating q, we can use the equation q = ΔH to find the enthalpy change of the reaction (ΔH). Since no heat is lost to the surroundings or the coffee cup, the amount of heat transferred (q) is equal to the enthalpy change (ΔH) of the reaction.